AP Chemistry practice test on chemical equations and stoichiometry with questions on balancing equations, mole ratios, and mass calculations.

Let’s solve each problem step by step.

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Problem 1:
Balance: NH₃ + O₂ → NO₂ + H₂O
We need smallest whole-number coefficients.

Start with N: 1 N on left (NH₃), 1 N on right (NO₂) → OK.
H: 3 H in NH₃, 2 H in H₂O → need even number of H on left. Try 4 NH₃ → 12 H → need 6 H₂O.

So try:
4 NH₃ + ? O₂ → ? NO₂ + 6 H₂O

Now N: 4 NH₃ → 4 NO₂
H: 4×3 = 12 H → 6 H₂O = 12 H ✔️
O: Right side: 4 NO₂ has 8 O, 6 H₂O has 6 O → total 14 O atoms
Left: O₂ provides 2 O per molecule → need 7 O₂

So balanced equation:
4 NH₃ + 7 O₂ → 4 NO₂ + 6 H₂O

Now: 1.00 mol NH₃ requires how many mol O₂?
From equation: 4 mol NH₃ : 7 mol O₂
→ 1 mol NH₃ : 7/4 = 1.75 mol O₂

Answer: d) 1.75

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Problem 2:
Combustion of acetaldehyde, CH₃CHO.
Combustion means reaction with O₂ → CO₂ + H₂O.

CH₃CHO has: C=2, H=4, O=1
So: CH₃CHO + O₂ → 2 CO₂ + 2 H₂O
Check atoms:

Left: C=2, H=4, O=1 + ?
Right: C=2, H=4, O=2×2 + 2×1 = 4+2 = 6 O
So need 5 O from O₂ → 5/2 O₂

Balanced:
CH₃CHO + ⁵⁄₂ O₂ → 2 CO₂ + 2 H₂O
Multiply by 2 to avoid fractions:
2 CH₃CHO + 5 O₂ → 4 CO₂ + 4 H₂O

So for 1 mol CH₃CHO, need 5/2 = 2.5 mol O₂

Answer: c) 2.5

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Problem 3:
Balance: KClO₃ → KCl + O₂
K and Cl are balanced (1 each). O: 3 on left, 2 on right.

Try 2 KClO₃ → 2 KCl + 3 O₂
Check:
K: 2 = 2 ✔️
Cl: 2 = 2 ✔️
O: 2×3 = 6 → 3×2 = 6 ✔️

Coefficients: 2, 2, 3 → sum = 2 + 2 + 3 = 7

Answer: c) 7

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Problem 4:
Combustion of propane, C₃H₈
C₃H₈ + O₂ → CO₂ + H₂O

C: 3 → 3 CO₂
H: 8 → 4 H₂O
O needed: 3×2 + 4×1 = 6 + 4 = 10 O atoms → 5 O₂

Balanced: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

So 1 mol C₃H₈ requires 5 mol O₂

Answer: c) 5

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Problem 5:
16 g CH₄ burned with excess O₂. Find total mass of products.

Reaction: CH₄ + 2 O₂ → CO₂ + 2 H₂O
Molar mass CH₄ = 12 + 4×1 = 16 g/mol
So 16 g = 1 mol CH₄

From equation: 1 mol CH₄ → 1 mol CO₂ + 2 mol H₂O
Mass of CO₂ = 12 + 2×16 = 44 g
Mass of 2 H₂O = 2×(2+16) = 2×18 = 36 g
Total products = 44 + 36 = 80 g

Answer: a) 80 g

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Problem 6:
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
25 g Al reacts with excess HCl. Find mass of H₂ formed.

Molar mass Al = 27 g/mol
Moles Al = 25 / 27 ≈ 0.9259 mol

From equation: 2 mol Al → 3 mol H₂
So 0.9259 mol Al → (3/2) × 0.9259 = 1.3889 mol H₂

Molar mass H₂ = 2 g/mol
Mass H₂ = 1.3889 × 2 ≈ 2.778 g ≈ 2.8 g

Answer: d) 2.8 g

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Problem 7:
3 Fe + 2 O₂ → Fe₃O₄
6.00 g O₂ reacts with Fe. Find grams of Fe₃O₄ formed.

Molar mass O₂ = 32 g/mol
Moles O₂ = 6.00 / 32 = 0.1875 mol

From equation: 2 mol O₂ → 1 mol Fe₃O₄
So 0.1875 mol O₂ → 0.1875 / 2 = 0.09375 mol Fe₃O₄

Molar mass Fe₃O₄ = 3×55.85 + 4×16 = 167.55 + 64 = 231.55 g/mol
(But problem gives no atomic masses — maybe expects using Fe = 56, O = 16)

Using Fe = 56, O = 16:
Fe₃O₄ = 3×56 + 4×16 = 168 + 64 = 232 g/mol

Mass = 0.09375 × 232 =
0.09375 × 200 = 18.75
0.09375 × 32 = 3.0
Total = 21.75 g ≈ 21.7 g

Answer: d) 21.7

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Problem 8:
2 MnO₂ + 4 KOH + O₂ + Cl₂ → 2 KMnO₄ + 2 KCl + 2 H₂O
100 g of each reactant. Which is limiting?

Molar masses:
MnO₂ = 86.9 g/mol
KOH = 56.1 g/mol
O₂ = 32.0 g/mol
Cl₂ = 70.9 g/mol

Moles available:
- MnO₂: 100 / 86.9 ≈ 1.151 mol
- KOH: 100 / 56.1 ≈ 1.782 mol
- O₂: 100 / 32.0 = 3.125 mol
- Cl₂: 100 / 70.9 ≈ 1.410 mol

Stoichiometric ratios (from equation):
2 MnO₂ : 4 KOH : 1 O₂ : 1 Cl₂
Divide all by 2 → 1 MnO₂ : 2 KOH : 0.5 O₂ : 0.5 Cl₂

Now find how many “reaction units” each can support:

- MnO₂: 1.151 mol / 2 = 0.5755
- KOH: 1.782 / 4 = 0.4455
- O₂: 3.125 / 1 = 3.125
- Cl₂: 1.410 / 1 = 1.410

Smallest is KOH → limiting reagent is KOH

Answer: b) KOH

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Problem 9:
3 NO₂ + H₂O → 2 HNO₃ + NO
92.0 g NO₂ and 36.0 g H₂O → how many grams HNO₃?

Molar masses:
NO₂ = 14 + 32 = 46 g/mol
H₂O = 18 g/mol
HNO₃ = 1 + 14 + 48 = 63 g/mol

Moles:
- NO₂: 92.0 / 46 = 2.00 mol
- H₂O: 36.0 / 18 = 2.00 mol

From equation: 3 mol NO₂ : 1 mol H₂O
So required ratio = 3:1
We have 2 mol NO₂ and 2 mol H₂O → NO₂ is limiting? Let's check:

If all 2.00 mol NO₂ react, need (1/3)×2.00 = 0.6667 mol H₂O → we have 2.00 mol, so yes, NO₂ is limiting.

From 3 mol NO₂ → 2 mol HNO₃
So 2.00 mol NO₂ → (2/3)×2.00 = 1.333... mol HNO₃

Mass = 1.333... × 63 = (4/3) × 63 = 84 g

Answer: c) 84

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Final Answers Summary:
1. d
2. c
3. c
4. c
5. a
6. d
7. d
8. b
9. c

Final Answer:
1. d
2. c
3. c
4. c
5. a
6. d
7. d
8. b
9. c