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Chapter 5 · Free lesson

Stoichiometry and reaction amounts

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YOUR CHAPTER PLAN

What you will study

Balance equations before using the recipe
Five common reaction types
A balanced reaction is a recipe
Grams A ↔ moles A ↔ moles B ↔ grams B
The limiting reactant
Yield tells how close the result came

By the end, you can…

Balance a chemical equation using the smallest whole-number coefficients, and verify every element on both sides.
Classify a reaction as synthesis, decomposition, single displacement, double displacement, or combustion.
Use a balanced mole ratio to find product amount.
Identify the given substance A and requested substance B, choose the start and stop units, and use molar mass → coefficient ratio → molar mass to convert reaction amounts.
Identify the limiting reactant by product capacity.
Calculate percent yield.

These are the skills to check, not just pages to finish. Use the section goals while studying, then try the chapter checks without hints.

Read the idea first. Then notice the clue, follow the route, and make a choice you can explain. You can return to the worked example if a practice question feels unfamiliar.

5.1 Balance equations before using the recipe

A reaction rearranges atoms. It never creates or removes them. Balancing makes the count of each kind of atom equal on both sides. Here the formulas are given; you choose how many of each.

What you’ll use from earlier

Read subscripts and count atoms in a formula (Chapter 2). Change coefficients, never subscripts.

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

An unbalanced water reaction has two oxygen atoms on the left and one on the right. Two hydrogen molecules plus one oxygen molecule form two water molecules, conserving four hydrogen and two oxygen atoms. — 2H₂ + O₂ → 2H₂O: the same atoms, rearranged. Count each colored atom, not the molecules.

Two numbers with different jobs

Number	Where	Example	Meaning
Subscript	Small, inside the formula	H₂O	2 H and 1 O. Leave it alone.
Coefficient	In front of the formula	3H₂O	3 × 2 = 6 H and 3 × 1 = 3 O.
No coefficient	Nothing written	H₂O	Means 1.

Reactants sit left of the arrow; products sit right.

Fixing a count: the right way and the wrong way

Change a coefficient

H₂O → 2H₂O
Still water, just more of it.

Change a subscript

H₂O → H₂O₂
Now it is hydrogen peroxide, a different substance.

Order of attack

Start with the most complicated formula, the one with the most atoms or elements.
Next, balance elements that show up in only one formula on each side. Leave lone elements (O₂, H₂, Fe) for last. Changing their coefficient changes only that element, so nothing else is disturbed.
If a group like SO₄, PO₄, NO₃ or OH (a polyatomic ion) appears unchanged on both sides, count it as one unit.
Stuck with an odd number of O atoms? Use a fraction, such as 72O₂. Then multiply every coefficient by 2 to clear it.
At the end, recount every element on both sides.

Before you calculate

Use the smallest whole-number coefficients. A 1 can stay unwritten.
Count each element on both sides before checking an answer.
No moles or grams are needed yet. The next lesson turns coefficients into a mole recipe.

1 · Recognize the clue

Look for this:an equation with an arrow and the word “balance,” or coefficients asked for, so every element’s atom count matches on both sides.

2 · Follow the route

Count the atoms of each element on each side.
Start with the most complicated formula; leave lone elements (O₂, H₂, Fe) for last.
Change one coefficient at a time; it multiplies the whole formula.
Recount every element after each change.
Clear any fractions, reduce to the smallest whole numbers, then check again.

On the periodic table: The periodic table names each symbol (H, O, Al). Balancing needs only atom counts, not masses.

Worked example 1

Balance: H₂ + O₂ → H₂O

1
Count first
Oxygen does not match. Hydrogen matches for now.
Element Left Right
H 2 2
O 2 1
2
Put 2 before water
Two waters give 2 O, but also 4 H. Recount both elements.
H₂ + O₂ → 2H₂O
Element Left Right
H 2 4
O 2 2
3
Put 2 before hydrogen
Two H₂ give 4 H. Both sides now have 4 H and 2 O.
2H₂ + O₂ → 2H₂O
Element Left Right
H 2 × 2 = 4 2 × 2 = 4
O 1 × 2 = 2 2 × 1 = 2
✓
Answer
2H₂ + O₂ → 2H₂O
Check: H 4 = 4; O 2 = 2.

Element	Left	Right
H	2	2
O	2	1

Element	Left	Right
H	2	4
O	2	2

Element	Left	Right
H	2 × 2 = 4	2 × 2 = 4
O	1 × 2 = 2	2 × 1 = 2

Worked example 2

Balance: Al + O₂ → Al₂O₃

1
Compare oxygen counts
O₂ holds 2 O; Al₂O₃ holds 3 O. The smallest multiple of both 2 and 3 is 6.
O₂: 2 O | Al₂O₃: 3 O → shared total: 6 O
2
Make 6 O on each side
Put 3 before O₂ and 2 before Al₂O₃.
Al + 3O₂ → 2Al₂O₃
Left: 3 × 2 = 6 O Right: 2 × 3 = 6 O
3
Fix aluminum
The products now hold 2 × 2 = 4 Al, so put 4 before Al.
4Al + 3O₂ → 2Al₂O₃
Element Left Right
Al 4 2 × 2 = 4
O 3 × 2 = 6 2 × 3 = 6
✓
Answer
4Al + 3O₂ → 2Al₂O₃
Check: Al 4 = 2 × 2; O 3 × 2 = 2 × 3.

Element	Left	Right
Al	4	2 × 2 = 4
O	3 × 2 = 6	2 × 3 = 6

Worked example 3

Balance: CH₄ + O₂ → CO₂ + H₂O

1
Check carbon
1 C on each side already. Keep both coefficients at 1.
Element Left Right
C 1 1
H 4 2
O 2 2 + 1 = 3
2
Balance hydrogen
CH₄ has 4 H. Two waters give 2 × 2 = 4 H.
CH₄ + O₂ → CO₂ + 2H₂O
3
Count all product oxygen
Add oxygen from every product, not just one formula.
CO₂: 2 O + 2H₂O: 2 × 1 = 2 O = 4 O
4
Balance oxygen
Two O₂ give 4 O. Leaving oxygen until last avoids repeated changes.
CH₄ + 2O₂ → CO₂ + 2H₂O
Element Left Right
C 1 1
H 4 2 × 2 = 4
O 2 × 2 = 4 2 + 2 = 4
✓
Answer
CH₄ + 2O₂ → CO₂ + 2H₂O
Check: C 1 = 1; H 4 = 2 × 2; O 2 × 2 = 2 + 2.
Odd oxygen? Use a fraction
In C₂H₆ combustion, the products hold 2 × 2 + 3 × 1 = 7 O, an odd number. Use 72O₂, then multiply every coefficient by 2.
C₂H₆ + 72O₂
→ 2CO₂ + 3H₂O
× 2:
2C₂H₆ + 7O₂
→ 4CO₂ + 6H₂O
C: 2 × 2 = 4 = 4
H: 2 × 6 = 12 = 6 × 2 = 12
O: 7 × 2 = 14 = 4 × 2 + 6 = 14

Element	Left	Right
C	1	1
H	4	2
O	2	2 + 1 = 3

Element	Left	Right
C	1	1
H	4	2 × 2 = 4
O	2 × 2 = 4	2 + 2 = 4

Worked example 4

Balance: Ca(OH)₂ + H₃PO₄ → Ca₃(PO₄)₂ + H₂O

1
Count, keeping PO₄ together
PO₄ appears unchanged on both sides, so count it as one unit. “Other O” means O atoms outside PO₄.
Atom or unit Left Right
Ca 1 3
PO₄ 1 2
H 2 + 3 = 5 2
Other O 2 1
2
Start with Ca₃(PO₄)₂
It is the most complicated formula. It holds 2 PO₄ units, so put 2 before H₃PO₄.
Ca(OH)₂ + 2H₃PO₄
→ Ca₃(PO₄)₂ + H₂O
PO₄: 2 × 1 = 2 = 1 × 2 = 2
3
Balance calcium
Ca₃(PO₄)₂ holds 3 Ca, so put 3 before Ca(OH)₂.
3Ca(OH)₂ + 2H₃PO₄
→ Ca₃(PO₄)₂ + H₂O
Ca: 3 × 1 = 3 = 3
4
Balance H with water
Add the hydrogen from both reactants. Each water holds 2 H, so divide by 2.
H: 3 × 2 + 2 × 3 = 12
12 ÷ 2 = 6H₂O
3Ca(OH)₂ + 2H₃PO₄
→ Ca₃(PO₄)₂ + 6H₂O
5
Recount everything
The other O now matches too: 6 = 6.
Atom or unit Left Right
Ca 3 × 1 = 3 3
PO₄ 2 × 1 = 2 1 × 2 = 2
H 6 + 6 = 12 6 × 2 = 12
Other O 3 × 2 = 6 6 × 1 = 6
✓
Answer
3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O
Check single atoms: Ca 3 = 3; P 2 = 2; H 12 = 12; O 14 = 14.

Atom or unit	Left	Right
Ca	1	3
PO₄	1	2
H	2 + 3 = 5	2
Other O	2	1

Atom or unit	Left	Right
Ca	3 × 1 = 3	3
PO₄	2 × 1 = 2	1 × 2 = 2
H	6 + 6 = 12	6 × 2 = 12
Other O	3 × 2 = 6	6 × 1 = 6

Section practice · 10 questions · tap to open

What this set is for: practicing the method from this section while it’s fresh. Every problem uses only what this section has taught.

Balance 1

H₂ + Cl₂ → HCl

Give me a hint

Count the start: left has 2 H and 2 Cl; right has 1 H and 1 Cl.

Check the balanced equation and atom counts

Answer: H₂ + Cl₂ → 2HCl
That’s right: H: 2 = 2; Cl: 2 = 2.

Explain the worked problem to me

What you know: H₂ + Cl₂ → HCl. Left: 2 H, 2 Cl. Right: 1 H, 1 Cl. What you want: The smallest whole-number coefficients that make every count equal.

Both elements are short on the right by the same amount. Each HCl holds only 1 H and 1 Cl.
Put 2 before HCl. Now the right has 2 × 1 = 2 H and 2 × 1 = 2 Cl.
Recount the left: H₂ gives 2 H, Cl₂ gives 2 Cl. Everything matches: H₂ + Cl₂ → 2HCl.

Check: H: 2 = 2. Cl: 2 = 2. The coefficients 1, 1, 2 cannot be reduced further.

Watch out: Do not write H₂Cl₂ or HCl₂. Changing a subscript makes a different substance. Change coefficients only.

Count the start: left has 2 H and 2 Cl; right has 1 H and 1 Cl.
Put 2 before HCl. This multiplies both H and Cl by 2.
Check: both sides now have 2 H and 2 Cl. The coefficients 1, 1, 2 cannot be reduced.

Answer: H₂ + Cl₂ → 2HCl
H: 2 = 2; Cl: 2 = 2.

Balance 2

Na + Cl₂ → NaCl

Give me a hint

Cl₂ gives 2 Cl on the left, but NaCl has only 1 Cl on the right.

Check the balanced equation and atom counts

Answer: 2Na + Cl₂ → 2NaCl
That’s right: Na: 2 = 2; Cl: 2 = 2.

Explain the worked problem to me

What you know: Na + Cl₂ → NaCl. Left: 1 Na, 2 Cl. Right: 1 Na, 1 Cl. What you want: The smallest whole-number coefficients.

Fix chlorine first. Cl₂ brings 2 Cl, so put 2 before NaCl: 2 × 1 = 2 Cl on the right.
That 2 also doubles sodium on the right: 2 × 1 = 2 Na. The left has only 1 Na.
Put 2 before Na. Now 2Na + Cl₂ → 2NaCl.

Check: Na: 2 = 2. Cl: 2 = 2. Recount every element after each change, as you just did.

Watch out: After fixing chlorine, students often forget that the coefficient on NaCl also changed the sodium count.

Cl₂ gives 2 Cl on the left, but NaCl has only 1 Cl on the right.
Put 2 before NaCl to get 2 Cl. This also gives 2 Na on the right.
Put 2 before Na. Both elements match; the smallest coefficients are 2, 1, 2.

Answer: 2Na + Cl₂ → 2NaCl
Na: 2 = 2; Cl: 2 = 2.

Balance 3

Mg + O₂ → MgO

Give me a hint

O₂ has 2 O, but MgO has only 1 O.

Check the balanced equation and atom counts

Answer: 2Mg + O₂ → 2MgO
That’s right: Mg: 2 = 2; O: 2 = 2.

Explain the worked problem to me

What you know: Mg + O₂ → MgO. Left: 1 Mg, 2 O. Right: 1 Mg, 1 O. What you want: The smallest whole-number coefficients.

Fix oxygen. O₂ has 2 O, so put 2 before MgO: 2 × 1 = 2 O on the right.
Recount: 2MgO also holds 2 × 1 = 2 Mg. The left has 1 Mg.
Put 2 before Mg. Result: 2Mg + O₂ → 2MgO.

Check: Mg: 2 = 2. O: 2 = 2. The set 2, 1, 2 has no common factor, so it is smallest.

Watch out: Never write MgO₂ to fix oxygen. That changes a subscript and the substance.

O₂ has 2 O, but MgO has only 1 O.
Put 2 before MgO. The right now has 2 O and 2 Mg.
Put 2 before Mg. Check both elements: 2 Mg and 2 O on each side.

Answer: 2Mg + O₂ → 2MgO
Mg: 2 = 2; O: 2 = 2.

Balance 4

N₂ + H₂ → NH₃

Give me a hint

N₂ has 2 N. Put 2 before NH₃ to make 2 N on the right.

Check the balanced equation and atom counts

Answer: N₂ + 3H₂ → 2NH₃
That’s right: N: 2 = 2; H: 6 = 6.

Explain the worked problem to me

What you know: N₂ + H₂ → NH₃. Left: 2 N, 2 H. Right: 1 N, 3 H. What you want: The smallest whole-number coefficients.

Fix nitrogen. Put 2 before NH₃: 2 × 1 = 2 N on the right.
Recount hydrogen on the right: 2 × 3 = 6 H. The left has only 2 H.
Put 3 before H₂: 3 × 2 = 6 H. Result: N₂ + 3H₂ → 2NH₃.

Check: N: 2 = 2. H: 3 × 2 = 6 on the left; 2 × 3 = 6 on the right.

Watch out: A coefficient multiplies every atom in the formula. 2NH₃ has 6 H, not 3 or 5.

N₂ has 2 N. Put 2 before NH₃ to make 2 N on the right.
Those two NH₃ units contain 2 × 3 = 6 H.
Put 3 before H₂ to supply 3 × 2 = 6 H. Check: 2 N and 6 H on each side.

Answer: N₂ + 3H₂ → 2NH₃
N: 2 = 2; H: 6 = 6.

Balance 5

Al + Cl₂ → AlCl₃

Give me a hint

Chlorine comes in groups of 2 in Cl₂ and groups of 3 in AlCl₃. Use 6 as their first shared total.

Check the balanced equation and atom counts

Answer: 2Al + 3Cl₂ → 2AlCl₃
That’s right: Al: 2 = 2; Cl: 6 = 6.

Explain the worked problem to me

What you know: Al + Cl₂ → AlCl₃. Chlorine comes in groups of 2 on the left and 3 on the right. What you want: The smallest whole-number coefficients.

Find the smallest total that is a multiple of both 2 and 3: that is 6 Cl.
Make 6 Cl on each side: 3Cl₂ gives 3 × 2 = 6, and 2AlCl₃ gives 2 × 3 = 6.
Now the right has 2 × 1 = 2 Al. Put 2 before Al. Result: 2Al + 3Cl₂ → 2AlCl₃.

Check: Al: 2 = 2. Cl: 6 = 6.

Watch out: Aluminum looks balanced at the start (1 = 1), but it changes once you fix chlorine. Fix the 2-and-3 element first, then recount aluminum.

Chlorine comes in groups of 2 in Cl₂ and groups of 3 in AlCl₃. Use 6 as their first shared total.
Put 3 before Cl₂ and 2 before AlCl₃. Both sides now have 6 Cl.
The products have 2 Al. Put 2 before Al. Check: 2 Al and 6 Cl on each side.

Answer: 2Al + 3Cl₂ → 2AlCl₃
Al: 2 = 2; Cl: 6 = 6.

Balance 6

Fe + O₂ → Fe₂O₃

Give me a hint

O₂ has 2 O per molecule; Fe₂O₃ has 3. Their smallest shared oxygen total is 6.

Check the balanced equation and atom counts

Answer: 4Fe + 3O₂ → 2Fe₂O₃
That’s right: Fe: 4 = 4; O: 6 = 6.

Explain the worked problem to me

What you know: Fe + O₂ → Fe₂O₃. Oxygen comes in groups of 2 on the left and 3 on the right. What you want: The smallest whole-number coefficients.

The smallest shared oxygen total is 6. Use 3O₂ (3 × 2 = 6 O) and 2Fe₂O₃ (2 × 3 = 6 O).
Count iron on the right: 2 × 2 = 4 Fe.
Put 4 before Fe. Result: 4Fe + 3O₂ → 2Fe₂O₃.

Check: Fe: 4 = 2 × 2. O: 3 × 2 = 6 on the left; 2 × 3 = 6 on the right.

Watch out: The subscript 2 in Fe₂O₃ means each unit holds 2 Fe. Two units hold 4 Fe, not 2.

O₂ has 2 O per molecule; Fe₂O₃ has 3. Their smallest shared oxygen total is 6.
Put 3 before O₂ and 2 before Fe₂O₃. Now oxygen matches.
Two Fe₂O₃ units contain 4 Fe. Put 4 before Fe. Check: 4 Fe and 6 O on each side.

Answer: 4Fe + 3O₂ → 2Fe₂O₃
Fe: 4 = 4; O: 6 = 6.

Balance 7

H₂O₂ → H₂O + O₂

Give me a hint

Start with hydrogen: H₂O₂ and H₂O each have 2 H. Oxygen is unequal: 2 on the left and 3 total on the right.

Check the balanced equation and atom counts

Answer: 2H₂O₂ → 2H₂O + O₂
That’s right: H: 4 = 4; O: 4 = 2 + 2.

Explain the worked problem to me

What you know: H₂O₂ → H₂O + O₂. H: 2 = 2. O: 2 on the left, 1 + 2 = 3 on the right. What you want: The smallest whole-number coefficients.

Oxygen is uneven: 2 on the left, 3 on the right. Try 2H₂O₂, which gives the left 2 × 2 = 4 H and 2 × 2 = 4 O.
Match hydrogen with 2H₂O: 2 × 2 = 4 H on the right.
Count all product oxygen: 2H₂O gives 2 O, O₂ gives 2 O, total 4. It matches. Result: 2H₂O₂ → 2H₂O + O₂.

Check: H: 4 = 4. O: 4 = 2 + 2.

Watch out: Add oxygen from every product. Counting only the O₂ or only the water gives the wrong total.

Start with hydrogen: H₂O₂ and H₂O each have 2 H. Oxygen is unequal: 2 on the left and 3 total on the right.
Try 2H₂O₂ → 2H₂O + O₂. Doubling peroxide and water keeps hydrogen matched.
Check oxygen: 2 × 2 = 4 on the left; 2 × 1 + 2 = 4 on the right. Hydrogen is 4 on each side.

Answer: 2H₂O₂ → 2H₂O + O₂
H: 4 = 4; O: 4 = 2 + 2.

Balance 8

C₂H₄ + O₂ → CO₂ + H₂O

Give me a hint

C₂H₄ has 2 C. Put 2 before CO₂ to match carbon.

Check the balanced equation and atom counts

Answer: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
That’s right: C: 2 = 2; H: 4 = 4; O: 6 = 4 + 2.

Explain the worked problem to me

What you know: C₂H₄ + O₂ → CO₂ + H₂O. This is a combustion, so balance C, then H, then O. What you want: The smallest whole-number coefficients.

Carbon: C₂H₄ has 2 C. Put 2 before CO₂.
Hydrogen: C₂H₄ has 4 H. Put 2 before H₂O: 2 × 2 = 4 H.
Oxygen last: 2CO₂ gives 4 O, 2H₂O gives 2 O, total 6. Use 3O₂ (3 × 2 = 6). Result: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O.

Check: C: 2 = 2. H: 4 = 4. O: 6 = 4 + 2.

Watch out: Leave O₂ for last. It is the only formula with just one element, so changing it disturbs nothing else.

C₂H₄ has 2 C. Put 2 before CO₂ to match carbon.
C₂H₄ has 4 H. Put 2 before H₂O to match hydrogen.
Count product oxygen: 2CO₂ gives 4 O and 2H₂O gives 2 O, totaling 6. Put 3 before O₂. Check C, H, and O.

Answer: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
C: 2 = 2; H: 4 = 4; O: 6 = 4 + 2.

Balance 9

NaOH + H₂SO₄ → Na₂SO₄ + H₂O

Give me a hint

Na₂SO₄ contains 2 Na. Put 2 before NaOH.

Check the balanced equation and atom counts

Answer: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
That’s right: Na: 2 = 2; H: 4 = 4; S: 1 = 1; O: 6 = 6. You may track unchanged SO₄ as one group, then verify individual atoms.

Explain the worked problem to me

What you know: NaOH + H₂SO₄ → Na₂SO₄ + H₂O. The SO₄ group appears unchanged on both sides. What you want: The smallest whole-number coefficients.

Sodium: Na₂SO₄ has 2 Na. Put 2 before NaOH.
Hydrogen: left now has 2 (from 2NaOH) + 2 (from H₂SO₄) = 4 H. Put 2 before H₂O: 2 × 2 = 4 H.
Track SO₄ as one group: 1 = 1. Result: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

Check: Now verify single atoms. O: 2 + 4 = 6 on the left; 4 + 2 = 6 on the right. Na 2 = 2, H 4 = 4, S 1 = 1.

Watch out: Hydrogen sits in two reactants. Add both sources before choosing the water coefficient.

Na₂SO₄ contains 2 Na. Put 2 before NaOH.
The left now has 2 H from 2NaOH plus 2 H from H₂SO₄, totaling 4 H. Put 2 before H₂O.
Check sulfur: 1 on each side. Check oxygen: 2 + 4 = 6 on the left and 4 + 2 = 6 on the right. Na and H also match.

Answer: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Na: 2 = 2; H: 4 = 4; S: 1 = 1; O: 6 = 6. You may track unchanged SO₄ as one group, then verify individual atoms.

Balance 10

Ca(OH)₂ + HCl → CaCl₂ + H₂O

Give me a hint

Read the parentheses first: Ca(OH)₂ contains 1 Ca, 2 O, and 2 H.

Check the balanced equation and atom counts

Answer: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
That’s right: Ca: 1 = 1; O: 2 = 2; H: 4 = 4; Cl: 2 = 2. The parentheses give 2 O and 2 H in Ca(OH)₂.

Explain the worked problem to me

What you know: Ca(OH)₂ + HCl → CaCl₂ + H₂O. Read the parentheses: Ca(OH)₂ has 1 Ca, 2 O and 2 H. What you want: The smallest whole-number coefficients.

Chlorine: CaCl₂ has 2 Cl. Put 2 before HCl.
Hydrogen on the left: 2 (from OH groups) + 2 (from 2HCl) = 4 H. Put 2 before H₂O: 2 × 2 = 4 H.
Oxygen: 2 on the left; 2 × 1 = 2 on the right. Result: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O.

Check: Ca: 1 = 1. O: 2 = 2. H: 4 = 4. Cl: 2 = 2.

Watch out: The 2 after the parentheses multiplies both O and H. Ca(OH)₂ has 2 H, not 1.

Read the parentheses first: Ca(OH)₂ contains 1 Ca, 2 O, and 2 H.
CaCl₂ needs 2 Cl. Put 2 before HCl. This supplies 2 more H, giving 4 H total on the left.
Put 2 before H₂O to get 4 H and 2 O on the right. Check: Ca 1, Cl 2, H 4, and O 2 on each side.

Answer: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
Ca: 1 = 1; O: 2 = 2; H: 4 = 4; Cl: 2 = 2. The parentheses give 2 O and 2 H in Ca(OH)₂.

5.1a Five common reaction types

Most reactions follow one of five patterns. Naming the pattern tells you what kind of change happened, and later helps you predict products.

What you’ll use from earlier

Balanced equations (5.1).
Ionic formulas (2.3).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

Five patterns at a glance

Letters stand for elements or ion groups. Watch which partners stay together and which swap.

The five types

Type	Pattern	Example
Synthesis (combination)	A + B → AB	H₂ + Cl₂ → 2HCl
Decomposition	AB → A + B	2H₂O₂ → 2H₂O + O₂
Single displacement	A + BC → AC + B	Zn + 2HCl → ZnCl₂ + H₂
Double displacement	AB + CD → AD + CB	AgNO₃ + NaCl → AgCl + NaNO₃
Combustion	fuel + O₂ → …	CH₄ + 2O₂ → CO₂ + 2H₂O

Synthesis: several substances make one. Decomposition: one breaks into several. Single displacement: a lone element replaces one in a compound. Double displacement: two ionic compounds swap partners. Combustion: a substance reacts with O₂, usually releasing heat and light.

Double displacement: in AgNO3 plus NaCl, the ions swap partners to form AgCl and NaNO3. — Double displacement: Ag⁺ and Na⁺ trade partners. Section 6.3 shows how to predict whether a solid forms.

Clues to look for

Count the reactants and the products.
Look for O₂ as a reactant.
Look for an element standing alone.
A compound of only C and H that burns completely gives CO₂ and H₂O.

1 · Recognize the clue

Look for this:“classify” or a reaction type question; compare how many reactants and products there are, and whether an element stands alone.

2 · Follow the route

Count reactants and products: many to one, or one to many?
Check for O₂ as a reactant (combustion).
Check for a lone element replacing another (single displacement).
Check for two ionic compounds swapping partners (double displacement).

On the periodic table: The periodic table shows whether a lone element is a metal; in Fe + CuSO₄, one metal (Fe) replaces another (Cu).

Worked example

Classify each reaction: 2Na + Cl₂ → 2NaCl; CaCO₃ → CaO + CO₂; Fe + CuSO₄ → FeSO₄ + Cu; C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

1
Reaction 1: count partners
Two reactants make one product: synthesis.
2Na + Cl₂ → 2NaCl
Reactants Products Clue
2 1 many → one
2
Reaction 2: count partners
One reactant breaks into two products: decomposition.
CaCO₃ → CaO + CO₂
Reactants Products Clue
1 2 one → many
3
Reaction 3: look for a lone element
Iron, standing alone, replaces copper in the compound: single displacement.
Fe + CuSO₄ → FeSO₄ + Cu
Reactants Products Clue
Fe alone Cu alone lone element swaps in
4
Reaction 4: look for O₂
A carbon–hydrogen compound reacts with O₂ to make CO₂ and H₂O: combustion.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Reactants Products Clue
C₃H₈ + O₂ CO₂ + H₂O burns in O₂
✓
Answer
Synthesis, decomposition, single displacement, combustion.

Reactants	Products	Clue
2	1	many → one

Reactants	Products	Clue
1	2	one → many

Reactants	Products	Clue
Fe alone	Cu alone	lone element swaps in

Reactants	Products	Clue
C₃H₈ + O₂	CO₂ + H₂O	burns in O₂

Section practice · 5 questions · tap to open

What this set is for: practicing the method from this section while it’s fresh. Every problem uses only what this section has taught.

Practice 1

Classify: 2KClO₃ → 2KCl + 3O₂

Give me a hint

How many reactants? How many products?

Check the answer

Answer: Decomposition

Explain the worked problem to me

What you know: 2KClO₃ → 2KCl + 3O₂. What you want: The reaction type.

Count the reactants: only one substance, KClO₃.
Count the products: two substances, KCl and O₂.
One substance breaking into two or more fits AB → A + B. This is Decomposition.

Why it matters: Classifying a reaction tells you what kind of products to expect before you balance anything.

Watch out: O₂ appears, but as a product. Combustion needs O₂ as a reactant.

That’s right: Decomposition

Practice 2

Classify: Mg + 2HCl → MgCl₂ + H₂

Give me a hint

Look for an element standing alone on each side.

Check the answer

Answer: Single displacement

Explain the worked problem to me

What you know: Mg + 2HCl → MgCl₂ + H₂. What you want: The reaction type.

Look for an element standing alone: Mg on the left, H₂ on the right.
Mg ends up in the compound with Cl. Hydrogen leaves the compound and stands alone.
An element replacing another in a compound fits A + BC → AC + B. This is Single displacement.

Why it matters: The lone element on each side is the fingerprint. Double displacement has compounds only.

Watch out: Two reactants and two products also describes double displacement. Check whether one reactant is a lone element.

That’s right: Single displacement

Practice 3

Classify: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

Give me a hint

Do two compounds trade partners?

Check the answer

Answer: Double displacement

Explain the worked problem to me

What you know: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl. What you want: The reaction type.

Both reactants are ionic compounds. No element stands alone.
Follow the ions: Ba²⁺ leaves Cl⁻ and pairs with SO₄²⁻. Na⁺ pairs with Cl⁻.
Two compounds swapping partners fits AB + CD → AD + CB. This is Double displacement.

Why it matters: Spotting a swap lets you predict products later, in Section 6.3, where you check whether a solid forms.

That’s right: Double displacement

Practice 4

Classify: N₂ + 3H₂ → 2NH₃

Give me a hint

Count the products.

Check the answer

Answer: Synthesis

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃. What you want: The reaction type.

Count the reactants: two substances, N₂ and H₂.
Count the products: one substance, NH₃. The coefficient 2 is still one kind of substance.
Two substances making one product fits A + B → AB. This is Synthesis.

Why it matters: Count kinds of substances, not coefficients. 2NH₃ is one product.

Watch out: Do not read the coefficient 2 as two products.

That’s right: Synthesis

Practice 5

Ethane, C₂H₆, burns completely in O₂. Write the balanced equation.

Give me a hint

Complete combustion of a carbon–hydrogen compound gives CO₂ and H₂O. Balance C, then H, then O.

Check the answer

Answer: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Explain the worked problem to me

What you know: Ethane, C₂H₆, burns completely in O₂. Complete combustion of a carbon–hydrogen compound gives CO₂ and H₂O. What you want: The balanced equation.

Write the skeleton: C₂H₆ + O₂ → CO₂ + H₂O. Carbon: put 2 before CO₂.
Hydrogen: 6 H, so put 3 before H₂O (3 × 2 = 6).
Oxygen on the right: 2 × 2 + 3 = 7, an odd number. O₂ only supplies O in pairs, so double every coefficient: 2C₂H₆, 4CO₂, 6H₂O.
Now right-side oxygen is 4 × 2 + 6 = 14. Use 7O₂. Result: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

Check: C: 4 = 4. H: 12 = 12. O: 14 = 8 + 6.

Watch out: An odd oxygen total cannot come from whole O₂ molecules. Doubling fixes it.

That’s right: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

5.2 A balanced reaction is a recipe

A balanced equation is a recipe. Its coefficients say how many moles of each substance react and form. Use that ratio to get from the amount you have to the amount you want.

What you’ll use from earlier

Coefficients count complete formula units; balance atoms first (5.1).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

What the recipe says

Reactants are the starting substances; products are what they become.
Balanced means the same number of each kind of atom on both sides.
Change only coefficients, the numbers in front of formulas. A new subscript means a new substance.
Coefficients give particle and mole ratios.

Reading 2H₂ + O₂ → 2H₂O

	2H₂	+ O₂	→ 2H₂O
Molecules	2	1	2
Moles	2 mol	1 mol	2 mol
Grams	4.032 g	32.00 g	36.03 g

Moles follow 2 : 1 : 2. Grams do not, because each substance has its own molar mass.

Moles wanted = moles given × balanced ratio

mol B = mol A × coefficient Bcoefficient A

A is the given substance; B is the wanted one. Coefficients come from the balanced equation.

Element symbols help you check a balanced equation. H and O identify the atoms conserved in the reaction. 2H₂ + O₂ → 2H₂O has 4 H atoms and 2 O atoms on each side. The mole ratio comes from balanced coefficients, not atomic numbers. — 4 H and 2 O on each side. The mole ratio comes from coefficients, not atomic numbers.

1 · Recognize the clue

Look for this:a balanced equation with moles of one substance given and moles of another asked, often with “excess” of the other reactant.

2 · Follow the route

Balance the equation.
Convert the given amount to moles.
Multiply by the coefficient ratio: wanted on top, given below.

On the periodic table: H: atomic number 1, molar mass 1.008 g/mol. O: atomic number 8, molar mass 16.00 g/mol.

On the periodic table: Atomic number is the proton count. The decimal mass is an average, not an isotope’s mass number.

On the periodic table: Use supplied rounded masses when a problem gives them.

Worked example

For 2H₂ + O₂ → 2H₂O, how much water can 3.0 mol O₂ make with excess H₂?

Conversion factor for this route

From the balanced equation
2 mol H₂O1 mol O₂

Excess H₂ means oxygen sets the amount.

1
Read the balanced recipe
The coefficients compare mole amounts.
1 mol O₂ → 2 mol H₂O
2
Set the wanted/given ratio
Wanted substance on top, given substance underneath.
3.0 mol O₂ × 2 mol H₂O1 mol O₂
3
Calculate
The oxygen unit cancels, leaving moles of water.
3.0 ~~mol O₂~~ × 2 mol H₂O1 ~~mol O₂~~ = 6.0 mol H₂O
✓
Answer
6.0 mol H₂O. Two moles of water form per mole of oxygen.

5.3 Grams A ↔ moles A ↔ moles B ↔ grams B

A balanced equation links moles, not grams. To go from grams of one substance to grams of another, convert to moles, use the coefficient ratio, then convert back.

What you’ll use from earlier

Grams ↔ moles (Chapter 4), balanced coefficients (5.1), and unit cancellation (Math first).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

Grams of A connects to moles of A through molar mass of A. Moles of A connects to moles of B through balanced coefficients. Moles of B connects to grams of B through molar mass of B. Each connection can be reversed. — A is what you know. B is what you need. Only the middle bridge changes the substance.

Grams A → moles A → moles B → grams B

g B = g A × 1 mol Amolar mass A (g) × coefficient B mol Bcoefficient A mol A × molar mass B (g)1 mol B

Each factor cancels the unit before it. The unit left must be grams of B.

Molar mass connects grams and moles of the same substance. Balanced coefficients connect moles of different reacting substances. — Molar mass: same substance, grams ↔ moles. Coefficients: different substances, moles ↔ moles.

Choose your start and stop

Given → wanted	Bridges you use
mol A → mol B	coefficient ratio only
g A → mol B	molar mass A, then coefficient ratio
mol A → g B	coefficient ratio, then molar mass B
g A → g B	all three bridges

A can be a product and B a reactant: the same route finds the reactant needed for a requested product.
These are theoretical amounts. If two reactant amounts are given, first find the limiting one (5.4).
Product mass includes atoms from every reacting substance.

1 · Recognize the clue

Look for this:grams or moles of one substance in a reaction and grams or moles of another asked, with molar masses given.

2 · Follow the route

Label the given substance A and the wanted substance B.
Grams A → moles A with A’s molar mass.
Moles A → moles B with the coefficient ratio, wanted on top.
Moles B → grams B with B’s molar mass.
Check that the only unit left is the one requested.

On the periodic table: Molar mass is the decimal number in an element’s box: O has an average atomic mass of 16.00 u, so 16.00 g/mol.

On the periodic table: For a compound, add the box masses of every atom in its formula.

Worked example 1

Given 12.0 g Mg with oxygen in excess, find grams of MgO. Balanced equation: 2Mg + O₂ → 2MgO.

1
Find the molar masses
On the periodic table (rounded for this example): Mg 24.0 g/mol, O 16.0 g/mol.
MgO = 24.0 + 16.0 = 40.0 g/mol

Face each bridge

The unit to cancel goes below; the wanted unit goes above.

Bridge	Factor	Why it faces this way
g Mg → mol Mg	1 mol Mg24.0 g Mg	g Mg below cancels the given g Mg.
mol Mg → mol MgO	2 mol MgO2 mol Mg	Balanced coefficients: wanted above, current below.
mol MgO → g MgO	40.0 g MgO1 mol MgO	mol MgO below cancels; g MgO stays.

3
Multiply across, then divide
Multiply the tops, multiply the bottoms, then divide.
12.0 ~~g Mg~~ × 1 ~~mol Mg~~24.0 ~~g Mg~~ × 2 ~~mol MgO~~2 ~~mol Mg~~ × 40.0 g MgO1 ~~mol MgO~~ = 96048.0 = 20.0 g MgO
4
Check the units
Every unit cancels except g MgO, which matches the question.
Unit Cancelled by
g Mg 24.0 g Mg
mol Mg 2 mol Mg
mol MgO 1 mol MgO
g MgO kept
✓
Answer
20.0 g MgO. It exceeds 12.0 g because the product includes oxygen that joined the magnesium.

Unit	Cancelled by
g Mg	24.0 g Mg
mol Mg	2 mol Mg
mol MgO	1 mol MgO
g MgO	kept

Worked example 2

For 2H₂ + O₂ → 2H₂O, find the mass of water made from 4 g H₂ with excess O₂. Use approximate molar masses H₂ = 2 g/mol and H₂O = 18 g/mol.

Conversion factors for this route

g H₂ → mol H₂	mol H₂ → mol H₂O	mol H₂O → g H₂O
1 mol H₂2 g H₂	2 mol H₂O2 mol H₂	18 g H₂O1 mol H₂O

H₂ = 2 g/mol and H₂O = 18 g/mol are different factors; use each for its own substance.

1
Convert hydrogen grams to moles
Coefficients compare moles, so convert the mass first. On the periodic table: H is about 1 g/mol (1.008 in its box), so H₂ = 2 × 1 = 2 g/mol.
4 ~~g H₂~~ × 1 mol H₂2 ~~g H₂~~ = 2 mol H₂
2
Use the reaction ratio
The hydrogen-to-water mole ratio is 1 : 1.
2 ~~mol H₂~~ × 2 mol H₂O2 ~~mol H₂~~ = 2 mol H₂O
3
Convert product moles to grams
Use the product’s molar mass to reach the requested unit. On the periodic table: H about 1 g/mol and O about 16 g/mol (1.008 and 16.00), so H₂O = 2 × 1 + 16 = 18 g/mol.
2 ~~mol H₂O~~ × 18 g H₂O1 ~~mol H₂O~~ = 36 g H₂O
✓
Answer
36 g H₂O, using the supplied approximate data. Oxygen contributes the extra mass.

Section practice · 4 questions · tap to open

What this set is for: practicing the method from this section while it’s fresh. Every problem uses only what this section has taught.

For 2Mg + O₂ → 2MgO, given 0.500 mol Mg; find mol MgO. Which bridge?

Give me a hint

Write the start unit and the requested unit. If both are moles, only one bridge connects them.

Check the answer

Coefficient ratio only: 0.500 × 2/2 = 0.500 mol MgO.

Explain the worked problem to me

What you know: 0.500 mol Mg. Equation: 2Mg + O₂ → 2MgO. What you want: Moles MgO, and which bridge to use.

Start unit is moles A (Mg). Stop unit is moles B (MgO). Both are moles, so only the middle bridge is needed.
Use the coefficient ratio, wanted over given: 0.500 mol Mg × (2 mol MgO / 2 mol Mg).
mol Mg cancels. 0.500 × 2 ÷ 2 = 0.500 mol MgO.

Check: The ratio is 1:1, so the moles should not change. They don't.

Watch out: No molar mass is needed here. Adding one would wrongly move you to grams.

Coefficient ratio only: 0.500 × 2/2 = 0.500 mol MgO.

YOUR PERIODIC-TABLE REFERENCE

12MgMagnesium24.31Group 2 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

For 2Mg + O₂ → 2MgO, given 12.0 g Mg (Mg = 24.0 g/mol); find mol MgO. Where do you stop?

Give me a hint

Start at grams of Mg. Write the unit the question asks for; that unit tells you where the route stops.

Check the answer

Stop at moles B: 12.0 / 24.0 × 2/2 = 0.500 mol MgO.

On the periodic table: Magnesium (Mg) has an average atomic mass of 24.31 u (molar mass 24.31 g/mol), the decimal number in its box; this problem uses a simplified 24.0 g/mol (24.31 would round to 24.3).

Explain the worked problem to me

What you know: 12.0 g Mg, Mg = 24.0 g/mol. Equation: 2Mg + O₂ → 2MgO. What you want: Moles MgO, and where to stop.

Start at grams A. First bridge: 12.0 g Mg × (1 mol Mg / 24.0 g Mg) = 0.500 mol Mg. g Mg cancels.
Middle bridge: 0.500 mol Mg × (2 mol MgO / 2 mol Mg) = 0.500 mol MgO. mol Mg cancels.
The question asks for moles, so stop at moles B: 0.500 mol MgO.

Check: 12.0 is half of 24.0, so 0.500 mol makes sense. Three significant figures throughout.

Watch out: Do not continue to grams MgO. Stop where the requested unit is.

Stop at moles B: 12.0 / 24.0 × 2/2 = 0.500 mol MgO.

YOUR PERIODIC-TABLE REFERENCE

12MgMagnesium24.31Group 2 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

For 2Mg + O₂ → 2MgO, given 0.500 mol Mg (MgO = 40.0 g/mol); find grams MgO. Where do you start?

Give me a hint

Is the given amount in grams or in moles? That decides whether you need the first bridge.

Check the answer

Start at moles A: 0.500 × 2/2 × 40.0 = 20.0 g MgO.

On the periodic table: You can also build this from the table: magnesium (Mg) 24.0 + oxygen (O) 16.0 = 40.0 g/mol for MgO, using simplified values of the box masses 24.31 and 16.00.

Explain the worked problem to me

What you know: 0.500 mol Mg. MgO = 40.0 g/mol. Equation: 2Mg + O₂ → 2MgO. What you want: Grams MgO, and where to start.

The given amount is already in moles, so start at moles A. Skip the first bridge.
Middle bridge: 0.500 mol Mg × (2 mol MgO / 2 mol Mg) = 0.500 mol MgO.
Last bridge: 0.500 mol MgO × (40.0 g MgO / 1 mol MgO) = 20.0 g MgO. mol MgO cancels.

Check: Only g MgO is left, matching the ask. 0.500 is half of 1 mole, and 20.0 is half of 40.0.

Watch out: Do not divide 0.500 by 24.0. It is already moles, not grams.

On the periodic table: You can also build this from the table: magnesium (Mg) 24.0 + oxygen (O) 16.0 = 40.0 g/mol for MgO, using simplified values of the box masses 24.31 and 16.00.

Start at moles A: 0.500 × 2/2 × 40.0 = 20.0 g MgO.

YOUR PERIODIC-TABLE REFERENCE

12MgMagnesium24.31Group 2 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

For 2Mg + O₂ → 2MgO, you need 40.0 g MgO. Find grams Mg required with excess oxygen (Mg = 24.0 g/mol; MgO = 40.0 g/mol).

Give me a hint

Here the given substance A is the product MgO and B is the reactant Mg. Convert grams MgO with MgO’s own molar mass first.

Check the answer

A = MgO; B = Mg. 40.0 / 40.0 × 2/2 × 24.0 = 24.0 g Mg.

On the periodic table: Magnesium (Mg) is taken as 24.0 g/mol here (24.31 in its box, simplified). For MgO, add oxygen (O) at 16.0 g/mol (16.00 in its box): 24.0 + 16.0 = 40.0 g/mol.

Explain the worked problem to me

What you know: Need 40.0 g MgO, oxygen in excess. MgO = 40.0 g/mol, Mg = 24.0 g/mol. Equation: 2Mg + O₂ → 2MgO. What you want: Grams Mg required.

Here A is the product MgO and B is the reactant Mg. The route works in either direction.
First bridge: 40.0 g MgO × (1 mol MgO / 40.0 g MgO) = 1.00 mol MgO.
Middle bridge: 1.00 mol MgO × (2 mol Mg / 2 mol MgO) = 1.00 mol Mg.
Last bridge: 1.00 mol Mg × (24.0 g Mg / 1 mol Mg) = 24.0 g Mg.

Check: Less Mg than MgO makes sense, because MgO also contains oxygen's mass.

Watch out: Use each substance's own molar mass. Do not convert MgO grams with 24.0.

On the periodic table: Magnesium (Mg) is taken as 24.0 g/mol here (24.31 in its box, simplified). For MgO, add oxygen (O) at 16.0 g/mol (16.00 in its box): 24.0 + 16.0 = 40.0 g/mol.

A = MgO; B = Mg. 40.0 / 40.0 × 2/2 × 24.0 = 24.0 g Mg.

YOUR PERIODIC-TABLE REFERENCE

12MgMagnesium24.31Group 2 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

5.4 The limiting reactant

The limiting reactant runs out first and stops the product. Find it by asking how much product each reactant could make.

What you’ll use from earlier

Balanced mole ratios and mass conversions (5.2–5.3).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

The recipe decides: building bicycles

Part	Have	Needed per bicycle	Bicycles possible
Wheels	10	2	5
Frames	3	1	3

Only 3 bicycles: extra wheels cannot replace missing frames.

Product capacity = available moles × mole ratio

mol product = mol reactant × product coefficientreactant coefficient

The balanced equation is the recipe. Do this for each reactant, same product; the smaller result limits.

Limiting reactant: the recipe decides what runs out. Conserved atoms: 6 H and 2 O before and after. Molecular counts illustrate the same coefficient ratio used with moles. — 6 H and 2 O before and after. Molecules follow the same ratio as moles.

Four H₂ molecules and three O₂ molecules react by 2H₂ + O₂ → 2H₂O, making four H₂O molecules with one O₂ left over. There are 8 H atoms and 6 O atoms both before and after. — H₂ runs out first; one O₂ is left in excess.

That picture as a before / change / after table

	2H₂	+ O₂	→ 2H₂O
Before	4	3	0
Change	−4	−2	+4
After	0	1	4

Changes follow the 2 : 1 : 2 coefficients. Atoms: 8 H and 6 O before and after.

Molar masses prepare a limiting-reactant comparison. N₂ and H₂ have different molar masses because their atoms differ. If amounts are in grams, convert each reactant to moles before using N₂ + 3H₂ → 2NH₃. If the question already gives moles, skip the mass conversion; compare product capacities. — Grams? Convert to moles first. Moles? Skip to product capacities.

1 · Recognize the clue

Look for this:amounts of two reactants given together, asking which one “limits” or the maximum product that can form.

2 · Follow the route

Get both reactant amounts in moles.
Find the product each reactant could make.
The smaller amount is the maximum product; its reactant limits.

On the periodic table: Molar masses come from the periodic table: N 14.01, H 1.008 g/mol. Use supplied masses when given.

On the periodic table: Amounts already in moles need no periodic-table conversion.

Worked example

For N₂ + 3H₂ → 2NH₃, you have 2 mol N₂ and 3 mol H₂. Find the limiting reactant and maximum ammonia.

Conversion factors (same product for both)

From N₂	From H₂
2 mol NH₃1 mol N₂	2 mol NH₃3 mol H₂

Amounts are already in moles, so no mass conversion is needed.

1
Check nitrogen’s capacity
What N₂ could make with unlimited H₂.
2 ~~mol N₂~~ × 2 mol NH₃1 ~~mol N₂~~ = 4 mol NH₃
2
Check hydrogen’s capacity
Same product, so the results compare fairly.
3 ~~mol H₂~~ × 2 mol NH₃3 ~~mol H₂~~ = 2 mol NH₃
3
Choose the smaller capacity
Hydrogen runs out first; nitrogen’s 4 mol is never reached.
Reactant Could make Result
N₂ 4 mol NH₃ excess
H₂ 2 mol NH₃ limiting
4
State both answers
The smaller capacity gives both answers.
Before / change / after (mol)
N₂ + 3H₂ → 2NH₃
Before 2 3 0
Change −1 −3 +2
After 1 0 2
H₂ reaches 0; 1 mol N₂ is left over.
✓
Answer
Limiting reactant: H₂. Maximum product: 2 mol NH₃.
Check it
More moles of H₂ (3) than N₂ (2), yet H₂ limits. The recipe decides, not raw counts.

Reactant	Could make	Result
N₂	4 mol NH₃	excess
H₂	2 mol NH₃	limiting

	N₂	+ 3H₂	→ 2NH₃
Before	2	3	0
Change	−1	−3	+2
After	1	0	2

5.5 Yield tells how close the result came

Real reactions rarely make the full predicted amount. Percent yield compares what you collected with what the limiting reactant could make.

What you’ll use from earlier

Theoretical yield from the reaction route (5.3–5.4); percentages (1.4 and 4.3).

Need a refresher? Open the table of contents and revisit that topic before starting.

The idea

Two yields

Theoretical yield

The maximum product, calculated from the limiting reactant.

Actual yield

The amount you collect, measured or supplied.

Percent yield

percent yield = actual yieldtheoretical yield × 100%

Use the same units for both amounts.

Theoretical yield can use table masses; percent yield cannot come from position. Add one N and three H contributions to find NH₃ molar mass. Use the balanced reaction to predict product mass, then actual ÷ theoretical × 100%. The actual recovered mass must be measured or supplied. The table does not give percent yield. — The table helps find theoretical yield. Actual yield must be measured or supplied.

1 · Recognize the clue

Look for this:“percent yield,” or an actual (collected) amount compared with a theoretical (predicted) amount.

2 · Follow the route

Find the actual and theoretical yields.
Put both in the same units.
Divide actual by theoretical, then multiply by 100%.

On the periodic table: Theoretical yield can use periodic-table molar masses, e.g. NH₃ = 14.01 + 3(1.008) = 17.03 g/mol.

On the periodic table: The table never gives actual or percent yield. In the example both masses are supplied, so no element box is needed.

Worked example

A reaction should make 25.0 g product but you collect 20.0 g. What is percent yield?

Collected vs predicted

1
Put actual above theoretical
What fraction of the predicted amount was collected?
20.0 g25.0 g
2
Divide
Matching gram units cancel, leaving a fraction.
20.0 g25.0 g = 0.800
3
Turn the fraction into percent
Multiply by 100 to express it per hundred.
0.800 × 100% = 80.0%
✓
Answer
80.0%. Actual is smaller than theoretical, so any answer above 100% is wrong here.

Mixed chapter review

Answer in your notebook before revealing help. Keep your first attempt separate from an answer corrected with help.

Mixed chapter review · 14 questions · tap to open

What this set is for: problems from every section of the chapter, mixed together and not labeled, the way a test asks them. Your job is to pick the right method yourself. Do this before the skill check.

Practice 1

For N₂ + 3H₂ → 2NH₃, how many moles of NH₃ can form from 7 mol N₂ with excess H₂?

Give me a hint

One reactant is stated to be in excess.

Check the answer

Answer: 14 mol

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 7 mol N₂; H₂ is in excess. What you want: Moles NH₃.

Excess H₂ means N₂ decides the amount. No limiting check needed.
Read the ratio, wanted over given: 2 mol NH₃ / 1 mol N₂.
7 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 14 mol NH₃.

Check: Each N₂ makes 2 NH₃, so the answer should be double 7.

Watch out: Do not flip the ratio to 1/2. The wanted substance goes on top.

Your route: Balanced coefficients → Wanted/given mole ratio

That’s right: 14 mol

YOUR PERIODIC-TABLE REFERENCE

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 2

For N₂ + 3H₂ → 2NH₃, how many moles of NH₃ can form from 8 mol N₂ with excess H₂?

Give me a hint

One reactant is stated to be in excess.

Check the answer

Answer: 16 mol

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 8 mol N₂; H₂ is in excess. What you want: Moles NH₃.

Excess H₂ means N₂ controls the product.
Ratio, wanted over given: 2 mol NH₃ / 1 mol N₂.
8 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 16 mol NH₃. mol N₂ cancels.

Check: Twice as many NH₃ as N₂: 16 is double 8.

Watch out: The 3 in front of H₂ plays no part here. Use only the two substances in the question.

Your route: Balanced coefficients → Wanted/given mole ratio

That’s right: 16 mol

YOUR PERIODIC-TABLE REFERENCE

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 3

In left-to-right order, which smallest whole-number coefficients balance Na + Cl₂ → NaCl?

Give me a hint

A reaction needs balancing.

Check the answer

Answer: 2, 1, 2

Explain the worked problem to me

What you know: Na + Cl₂ → NaCl. Left: 1 Na, 2 Cl. Right: 1 Na, 1 Cl. What you want: Coefficients in left-to-right order.

Fix chlorine: put 2 before NaCl, giving 2 Cl on the right.
That makes 2 Na on the right. Put 2 before Na.
Cl₂ keeps an unwritten 1. In order: 2, 1, 2.

Check: Na: 2 = 2. Cl: 2 = 2. The numbers share no common factor, so they are smallest.

Watch out: List the unwritten 1 for Cl₂. Skipping it gives "2, 2", which is incomplete.

Your route: Count each atom → Adjust coefficients only

That’s right: 2, 1, 2

Practice 4

For 2H₂ + O₂ → 2H₂O with excess oxygen, what water mass forms from 14 g H₂? Use H₂ = 2 and H₂O = 18 g/mol.

Give me a hint

The mass of one reactant is given; the product mass is requested.

Check the answer

Answer: 126 g

Explain the worked problem to me

What you know: 2H₂ + O₂ → 2H₂O; 14 g H₂; O₂ in excess. H₂ = 2 g/mol, H₂O = 18 g/mol. What you want: Grams H₂O.

Grams to moles: 14 g H₂ × (1 mol H₂ / 2 g H₂) = 7 mol H₂.
Reaction ratio: 7 mol H₂ × (2 mol H₂O / 2 mol H₂) = 7 mol H₂O.
Moles to grams: 7 mol H₂O × (18 g H₂O / 1 mol H₂O) = 126 g.

Check: Water outweighs the hydrogen because it includes the oxygen that joined. 126 − 14 = 112 g came from O₂.

Watch out: Do not multiply 14 by 18 directly. Grams of one substance never convert straight to grams of another.

Your route: Grams → Moles → Reaction ratio → Product grams

On the periodic table: You can also build these from the table: hydrogen (H) is about 1 g/mol and oxygen (O) about 16 g/mol (1.008 and 16.00 in their boxes), so H₂ = 2 × 1 = 2 g/mol and H₂O = 2 × 1 + 16 = 18 g/mol.

That’s right: 126 g

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 5

For 2H₂ + O₂ → 2H₂O with excess oxygen, what water mass forms from 16 g H₂? Use H₂ = 2 and H₂O = 18 g/mol.

Give me a hint

The mass of one reactant is given; the product mass is requested.

Check the answer

Answer: 144 g

Explain the worked problem to me

What you know: 2H₂ + O₂ → 2H₂O; 16 g H₂; O₂ in excess. H₂ = 2 g/mol, H₂O = 18 g/mol. What you want: Grams H₂O.

Grams to moles: 16 g H₂ × (1 mol H₂ / 2 g H₂) = 8 mol H₂.
Reaction ratio: 8 mol H₂ × (2 mol H₂O / 2 mol H₂) = 8 mol H₂O.
Moles to grams: 8 mol H₂O × (18 g H₂O / 1 mol H₂O) = 144 g.

Check: Reverse it: 144 ÷ 18 = 8 mol water, which needs 8 mol H₂ = 16 g. It matches.

Watch out: Use 2 g/mol for H₂, not 1. Hydrogen gas has two atoms per molecule.

Your route: Grams → Moles → Reaction ratio → Product grams

That’s right: 144 g

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 6

For N₂ + 3H₂ → 2NH₃, a mixture has 8 mol N₂ and 18 mol H₂. Which reactant limits the product?

Give me a hint

Amounts of both reactants are provided.

Check the answer

Answer: H₂

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 8 mol N₂ and 18 mol H₂. What you want: Which reactant limits.

N₂ capacity: 8 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 16 mol NH₃.
H₂ capacity: 18 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 12 mol NH₃.
12 mol is less than 16 mol. Hydrogen runs out first, so H₂ limits.

Check: 8 mol N₂ would need 8 × 3 = 24 mol H₂. Only 18 is there, so H₂ is short.

Watch out: Do not pick N₂ because 8 is smaller than 18. Compare product capacities, not starting amounts.

Your route: Product capacity for both → Smaller limits

That’s right: H₂

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 7

For N₂ + 3H₂ → 2NH₃, a mixture has 9 mol N₂ and 21 mol H₂. Which reactant limits the product?

Give me a hint

Amounts of both reactants are provided.

Check the answer

Answer: H₂

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 9 mol N₂ and 21 mol H₂. What you want: Which reactant limits.

N₂ capacity: 9 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 18 mol NH₃.
H₂ capacity: 21 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 14 mol NH₃.
14 mol is less than 18 mol. Hydrogen runs out first, so H₂ limits.

Check: 9 mol N₂ would need 9 × 3 = 27 mol H₂. Only 21 is available.

Watch out: The larger starting number (21) is still the limiting one. The recipe needs 3 H₂ per N₂.

Your route: Product capacity for both → Smaller limits

That’s right: H₂

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 8

For N₂ + 3H₂ → 2NH₃, 8 mol N₂ and 18 mol H₂ are available. What maximum NH₃ amount can form?

Give me a hint

Both reactants are given and a maximum product is requested.

Check the answer

Answer: 12 mol

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 8 mol N₂ and 18 mol H₂. What you want: Maximum moles NH₃.

Find the limiting reactant by product capacity. N₂: 8 × 2/1 = 16 mol NH₃. H₂: 18 × 2/3 = 12 mol NH₃.
H₂ gives the smaller capacity, so H₂ limits.
The limiting reactant's capacity is the maximum: 18 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 12 mol NH₃.

Check: 12 mol NH₃ uses 6 mol N₂, leaving 2 mol N₂ in excess.

Watch out: Do not report 16 mol. The excess reactant's capacity can never be reached.

Your route: Find limiting reactant → Its moles × Product/reactant ratio

That’s right: 12 mol

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 9

For N₂ + 3H₂ → 2NH₃, 9 mol N₂ and 21 mol H₂ are available. What maximum NH₃ amount can form?

Give me a hint

Both reactants are given and a maximum product is requested.

Check the answer

Answer: 14 mol

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 9 mol N₂ and 21 mol H₂. What you want: Maximum moles NH₃.

Product capacities: N₂: 9 × 2/1 = 18 mol NH₃. H₂: 21 × 2/3 = 14 mol NH₃.
The smaller capacity comes from H₂, so H₂ limits.
Maximum product: 21 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 14 mol NH₃.

Check: 14 mol NH₃ uses 7 mol N₂, leaving 2 mol N₂ unused.

Watch out: Do not report 18 mol, the excess reactant's capacity.

Your route: Find limiting reactant → Its moles × Product/reactant ratio

That’s right: 14 mol

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 10

Theoretical yield is 55 g and actual yield is 44 g. What is percent yield?

Give me a hint

Actual and theoretical yields are both given.

Check the answer

Answer: 80%

Explain the worked problem to me

What you know: Theoretical yield 55 g. Actual yield 44 g. What you want: Percent yield.

Put actual above theoretical: 44 g / 55 g. The gram units cancel.
Divide: 44 ÷ 55 = 0.80.
Turn it into a percent: 0.80 × 100% = 80%.

Check: Actual is less than theoretical, so the answer must be below 100%. It is.

Watch out: Do not divide 55 by 44. That gives 125%, which is impossible here.

Your route: Actual ÷ Theoretical × 100

That’s right: 80%

Practice 11

Theoretical yield is 60 g and actual yield is 48 g. What is percent yield?

Give me a hint

Actual and theoretical yields are both given.

Check the answer

Answer: 80%

Explain the worked problem to me

What you know: Theoretical yield 60 g. Actual yield 48 g. What you want: Percent yield.

Put actual above theoretical: 48 g / 60 g. Units match, so they cancel.
Divide: 48 ÷ 60 = 0.80.
Multiply by 100%: 0.80 × 100% = 80%.

Check: Reverse it: 80% of 60 g is 0.80 × 60 = 48 g, the actual yield.

Watch out: Theoretical always goes on the bottom. Flipping gives 125%.

Your route: Actual ÷ Theoretical × 100

That’s right: 80%

Practice 12

A student wants to read a mole ratio from the unbalanced equation Na + Cl₂ → NaCl. What must happen first?

Give me a hint

The task asks how to use a reaction equation.

Check the answer

Answer: Balance atoms by changing coefficients

Explain the worked problem to me

What you know: Na + Cl₂ → NaCl is unbalanced. Left: 1 Na, 2 Cl. Right: 1 Na, 1 Cl. What you want: What to do before reading a mole ratio.

Mole ratios come from coefficients. Unbalanced coefficients give a false recipe.
So first Balance atoms by changing coefficients: 2Na + Cl₂ → 2NaCl.
Only now read the ratio, for example 2 mol NaCl / 1 mol Cl₂.

Why it matters: Using the unbalanced 1:1 ratio would predict half the NaCl that 1 mol Cl₂ really makes.

Watch out: Balance with coefficients, never subscripts. Writing NaCl₂ invents a different substance.

Your route: Balance → Read ratio

That’s right: Balance atoms by changing coefficients

Practice 13

For N₂ + 3H₂ → 2NH₃, how many H₂ moles are required for 7 mol N₂?

Give me a hint

One reactant amount is given; the required amount of the other reactant is asked.

Check the answer

Answer: 21 mol

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 7 mol N₂. What you want: Moles H₂ required.

Both amounts are in moles, so only the coefficient ratio is needed.
Ratio, required over given: 3 mol H₂ / 1 mol N₂.
7 mol N₂ × (3 mol H₂ / 1 mol N₂) = 21 mol H₂.

Check: The recipe needs 3 H₂ per N₂. 21 is three times 7.

Watch out: Use 3/1, not 1/3. The substance you want goes on top.

Your route: Given moles → Required/given coefficient ratio

That’s right: 21 mol

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Practice 14

For N₂ + 3H₂ → 2NH₃, how many H₂ moles are required for 8 mol N₂?

Give me a hint

One reactant amount is given; the required amount of the other reactant is asked.

Check the answer

Answer: 24 mol

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 8 mol N₂. What you want: Moles H₂ required.

Moles to moles uses the coefficient ratio only.
Ratio, required over given: 3 mol H₂ / 1 mol N₂.
8 mol N₂ × (3 mol H₂ / 1 mol N₂) = 24 mol H₂. mol N₂ cancels.

Check: Reverse it: 24 ÷ 3 = 8 mol N₂. It matches.

Watch out: Do not use the NH₃ coefficient. Only N₂ and H₂ are in this question.

Your route: Given moles → Required/given coefficient ratio

That’s right: 24 mol

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill check: the chapter’s key skills

Try each fresh question without the worked examples. Record your first answer and whether you used a hint. Explain the route as well as the answer.

Skill check · 6 questions · tap to open

What this set is for: one question on each key skill from this chapter. Try each on your own, without hints, then mark how it went in the tracker below. A miss tells you which section to review.

Skill 1: Balance a chemical equation using the smallest whole-number coefficients, and verify every element on both sides.

Balance Al + O₂ → Al₂O₃ and explain your oxygen count.

Give me a hint

Count each element on both sides. Change coefficients only; recount after each change and reduce to the smallest whole numbers.

Check the answer

4Al + 3O₂ → 2Al₂O₃. Each side has 4 Al and 6 O atoms; coefficients are the smallest whole numbers.

Explain the worked problem to me

What you know: Al + O₂ → Al₂O₃. Oxygen comes in groups of 2 on the left and 3 on the right. What you want: The balanced equation and the oxygen count.

The smallest total that is a multiple of both 2 and 3 is 6 O.
Make 6 O on each side: 3O₂ gives 3 × 2 = 6; 2Al₂O₃ gives 2 × 3 = 6.
Aluminum on the right is now 2 × 2 = 4. Put 4 before Al. Result: 4Al + 3O₂ → 2Al₂O₃.

Check: Al: 4 = 4. O: 6 = 6. The set 4, 3, 2 has no common factor, so it is smallest.

Watch out: Count O₂ as 2 atoms per molecule. 3O₂ is 6 O atoms, not 3.

That’s right: 4Al + 3O₂ → 2Al₂O₃. Each side has 4 Al and 6 O atoms; coefficients are the smallest whole numbers.

YOUR PERIODIC-TABLE REFERENCE

8OOxygen16.00Group 16 · Period 2

13AlAluminum26.98Group 13 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 2: Classify a reaction as synthesis, decomposition, single displacement, double displacement, or combustion.

Classify: 2Al + 3Br₂ → 2AlBr₃

Give me a hint

Count the reactants and products.

Check the answer

Answer: Synthesis

Explain the worked problem to me

What you know: 2Al + 3Br₂ → 2AlBr₃. What you want: The reaction type.

Count reactants: two substances, Al and Br₂.
Count products: one substance, AlBr₃.
Two substances making one product fits A + B → AB. This is Synthesis.

Why it matters: Recognizing synthesis tells you the product is a single compound built from the reactants.

Watch out: Coefficients do not count as extra substances. 2AlBr₃ is still one product.

That’s right: Synthesis

Skill 3: Use a balanced mole ratio to find product amount.

For 2H₂ + O₂ → 2H₂O, find water from 1.5 mol O₂ with excess H₂.

Give me a hint

Use the wanted substance’s coefficient above the given substance’s coefficient. Cancel the given mole label.

Check the answer

1.5 × 2/1 = 3.0 mol H₂O.

Explain the worked problem to me

What you know: 2H₂ + O₂ → 2H₂O; 1.5 mol O₂; H₂ in excess. What you want: Moles H₂O.

Excess H₂ means O₂ controls the product.
Ratio, wanted over given: 2 mol H₂O / 1 mol O₂.
1.5 mol O₂ × (2 mol H₂O / 1 mol O₂) = 3.0 mol H₂O. Keep two significant figures, as in 1.5.

Check: Each O₂ makes 2 H₂O, so the answer should be double 1.5.

Watch out: Do not use the H₂ coefficient. Use only the given (O₂) and wanted (H₂O) substances.

That’s right: 1.5 × 2/1 = 3.0 mol H₂O.

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

8OOxygen16.00Group 16 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 4: Identify the given substance A and requested substance B, choose the start and stop units, and use molar mass → coefficient ratio → molar mass to convert reaction amounts.

For 2Mg + O₂ → 2MgO, find grams MgO from 12.0 g Mg with excess O₂. Use Mg = 24.0 g/mol and MgO = 40.0 g/mol.

Give me a hint

Convert the given grams to moles, apply the balanced coefficient ratio, then use the product’s molar mass to reach grams.

Check the answer

12.0 g Mg × (1 mol Mg / 24.0 g Mg) × (2 mol MgO / 2 mol Mg) × (40.0 g MgO / 1 mol MgO) = 20.0 g MgO. All units except g MgO cancel.

On the periodic table: You can also build these from the table: magnesium (Mg) is taken as 24.0 g/mol (24.31 in its box, simplified), and MgO = 24.0 + 16.0 for oxygen (O) = 40.0 g/mol.

Explain the worked problem to me

What you know: 2Mg + O₂ → 2MgO; 12.0 g Mg; O₂ in excess. Mg = 24.0 g/mol, MgO = 40.0 g/mol. What you want: Grams MgO.

Grams A → grams B uses all three bridges. First: 12.0 g Mg × (1 mol Mg / 24.0 g Mg) = 0.500 mol Mg.
Middle: 0.500 mol Mg × (2 mol MgO / 2 mol Mg) = 0.500 mol MgO.
Last: 0.500 mol MgO × (40.0 g MgO / 1 mol MgO) = 20.0 g MgO. Every unit except g MgO cancels.

Check: MgO weighs more than the Mg because it includes oxygen. 20.0 − 12.0 = 8.0 g O joined.

Watch out: Do not use 24.0 for both conversions. Each bridge uses its own substance's molar mass.

On the periodic table: You can also build these from the table: magnesium (Mg) is taken as 24.0 g/mol (24.31 in its box, simplified), and MgO = 24.0 + 16.0 for oxygen (O) = 40.0 g/mol.

That’s right: 12.0 g Mg × (1 mol Mg / 24.0 g Mg) × (2 mol MgO / 2 mol Mg) × (40.0 g MgO / 1 mol MgO) = 20.0 g MgO. All units except g MgO cancel.

YOUR PERIODIC-TABLE REFERENCE

12MgMagnesium24.31Group 2 · Period 3

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 5: Identify the limiting reactant by product capacity.

N₂ + 3H₂ → 2NH₃: 1 mol N₂ and 1 mol H₂. Which limits?

Give me a hint

Calculate the amount of the same product each reactant could make using the balanced mole ratio. The smaller capacity identifies the limiting reactant.

Check the answer

H₂: it can make only 2/3 mol NH₃; N₂ could make 2 mol.

Explain the worked problem to me

What you know: N₂ + 3H₂ → 2NH₃; 1 mol N₂ and 1 mol H₂. What you want: Which reactant limits.

N₂ capacity: 1 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 2 mol NH₃.
H₂ capacity: 1 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 2/3 mol NH₃.
2/3 mol is less than 2 mol, so hydrogen runs out first. H₂ limits: it can make only 2/3 mol NH₃.

Check: 1 mol N₂ would need 3 mol H₂. Only 1 mol is there.

Watch out: Equal starting moles do not mean equal use. The recipe uses 3 H₂ for each N₂.

That’s right: H₂: it can make only 2/3 mol NH₃; N₂ could make 2 mol.

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

7NNitrogen14.01Group 15 · Period 2

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Skill 6: Calculate percent yield.

Actual yield 9.0 g; theoretical 12 g. Percent?

Give me a hint

Divide actual yield by theoretical yield and multiply by 100. Use the same units in both amounts.

Check the answer

9.0/12 × 100 = 75%.

Explain the worked problem to me

What you know: Actual yield 9.0 g. Theoretical yield 12 g. What you want: Percent yield.

Put actual above theoretical: 9.0 g / 12 g. Same units, so they cancel.
Divide: 9.0 ÷ 12 = 0.75.
Multiply by 100%: 0.75 × 100% = 75%. Both values have two significant figures.

Check: Actual is below theoretical, so the result is under 100%. 75% of 12 g is 9.0 g.

Watch out: Dividing 12 by 9.0 gives 133%, a sign the fraction is upside down.

That’s right: 9.0/12 × 100 = 75%.

What to do next

Mark each skill: “I can do it independently,” “I needed help,” or “I need another explanation.” Revisit the section with the same goal for any skill needing help, then try a different question. A successful first attempt is evidence of progress; one answer alone does not establish lasting mastery.

Practice with instant feedback Next: Chapter 6

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Start the 20 free questions Get the full final · $7.99

Original ChemGem questions.

Stoichiometry and reaction amounts

What you will study

By the end, you can…

5.1 Balance equations before using the recipe

Balance: H2 + O2 → H2O

Balance: Al + O2 → Al2O3

Balance: CH4 + O2 → CO2 + H2O

Balance: Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O

Balance 1

Balance 2

Balance 3

Balance 4

Balance 5

Balance 6

Balance 7

Balance 8

Balance 9

Balance 10

5.1a Five common reaction types

Classify each reaction: 2Na + Cl2 → 2NaCl; CaCO3 → CaO + CO2; Fe + CuSO4 → FeSO4 + Cu; C3H8 + 5O2 → 3CO2 + 4H2O.

Practice 1

Practice 2

Practice 3

Practice 4

Practice 5

5.2 A balanced reaction is a recipe

For 2H2 + O2 → 2H2O, how much water can 3.0 mol O2 make with excess H2?

5.3 Grams A ↔ moles A ↔ moles B ↔ grams B

Given 12.0 g Mg with oxygen in excess, find grams of MgO. Balanced equation: 2Mg + O2 → 2MgO.

For 2H2 + O2 → 2H2O, find the mass of water made from 4 g H2 with excess O2. Use approximate molar masses H2 = 2 g/mol and H2O = 18 g/mol.

5.4 The limiting reactant

For N2 + 3H2 → 2NH3, you have 2 mol N2 and 3 mol H2. Find the limiting reactant and maximum ammonia.

5.5 Yield tells how close the result came

A reaction should make 25.0 g product but you collect 20.0 g. What is percent yield?

Mixed chapter review

Practice 1

Practice 2

Practice 3

Practice 4

Practice 5

Practice 6

Practice 7

Practice 8

Practice 9

Practice 10

Practice 11

Practice 12

Practice 13

Practice 14

Skill check: the chapter’s key skills

Skill 1: Balance a chemical equation using the smallest whole-number coefficients, and verify every element on both sides.

Skill 2: Classify a reaction as synthesis, decomposition, single displacement, double displacement, or combustion.

Skill 3: Use a balanced mole ratio to find product amount.

Skill 4: Identify the given substance A and requested substance B, choose the start and stop units, and use molar mass → coefficient ratio → molar mass to convert reaction amounts.

Skill 5: Identify the limiting reactant by product capacity.

Skill 6: Calculate percent yield.

What to do next

Take the 70-question practice final

Balance: H₂ + O₂ → H₂O

Balance: Al + O₂ → Al₂O₃

Balance: CH₄ + O₂ → CO₂ + H₂O

Balance: Ca(OH)₂ + H₃PO₄ → Ca₃(PO₄)₂ + H₂O

Classify each reaction: 2Na + Cl₂ → 2NaCl; CaCO₃ → CaO + CO₂; Fe + CuSO₄ → FeSO₄ + Cu; C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

For 2H₂ + O₂ → 2H₂O, how much water can 3.0 mol O₂ make with excess H₂?

Given 12.0 g Mg with oxygen in excess, find grams of MgO. Balanced equation: 2Mg + O₂ → 2MgO.

For 2H₂ + O₂ → 2H₂O, find the mass of water made from 4 g H₂ with excess O₂. Use approximate molar masses H₂ = 2 g/mol and H₂O = 18 g/mol.

For N₂ + 3H₂ → 2NH₃, you have 2 mol N₂ and 3 mol H₂. Find the limiting reactant and maximum ammonia.