Portage Learning CHEM 104 Module 3
Exam – Portage Learning – Academic
Year | Exam questions with verified
answers|| LATEST UPDATE 2026
CHEM 104 — MODULE 3 PRACTICE
QUESTIONS 2026
SECTION 1 — STOICHIOMETRY AND CHEMICAL
REACTIONS
1. How many moles are in 12 g of carbon?
A. 0.5 mol
B. 1 mol
C. 2 mol
D. 12 mol
Answer: B
Rationale: Molar mass of carbon = 12 g/mol → 12 g ÷ 12 g/mol = 1 mol
2. How many grams are in 2 moles of H₂O?
A. 36 g
B. 18 g
C. 9 g
D. 2 g
Answer: A
Rationale: Molar mass H₂O = 18 g/mol → 2 × 18 = 36 g
,3. What is the mass percent of O in H₂O?
A. 88.9%
B. 11.1%
C. 50%
D. 33%
Answer: A
Rationale: O mass = 16 g, H = 2 g → total = 18 g → %O = 16/18 × 100 ≈ 88.9%
4. Which is a balanced equation?
A. 2H₂ + O₂ → 2H₂O
B. H₂ + O₂ → H₂O
C. 2H₂ + O₂ → H₂O
D. H₂ + O → H₂O
Answer: A
Rationale: Both sides have 4 H atoms and 2 O atoms
5. How many moles of O₂ are needed to react with 4 moles of H₂ in 2H₂ + O₂ →
2H₂O?
A. 1 mole
B. 2 moles
C. 4 moles
D. 8 moles
Answer: A
Rationale: Ratio H₂:O₂ = 2:1 → 4 mol H₂ × (1/2) = 2 mol O₂ Wait—careful
Step: 2 H₂ + 1 O₂ → 2 H₂O
4 mol H₂ × (1 mol O₂ / 2 mol H₂) = 2 mol O₂ ✅
Correct Answer: B
6. In 2Na + Cl₂ → 2NaCl, the limiting reagent if 3 moles Na and 2 moles Cl₂ react
is:
A. Na
B. Cl₂
C. NaCl
, D. Both are in excess
Answer: B
Rationale: Ratio 2:1 → 3 mol Na needs 1.5 mol Cl₂ → 2 mol Cl₂ available → Cl₂ is in excess,
Na is limiting? Let's check:
2Na + Cl₂ → 2NaCl
3 mol Na × (1 mol Cl₂ / 2 mol Na) = 1.5 mol Cl₂ needed → 2 mol Cl₂ available → Na limits
reaction
✅ Limiting reagent = Na
7. How many grams of NaCl are produced from 1 mole Na + 1 mole Cl₂?
A. 58.44 g
B. 29.22 g
C. 117 g
D. 22.99 g
Answer: A
Rationale: 2Na + Cl₂ → 2NaCl → 1 mol Na reacts with 0.5 mol Cl₂ → forms 1 mol NaCl →
58.44 g
8. What is the empirical formula of a compound with 40% C, 6.7% H, 53.3% O?
A. CH₂O
B. C₂H₄O
C. CHO
D. C₂H₆O₂
Answer: A
Rationale:
C: 40 g → 40/12 ≈ 3.33 mol
H: 6.7 g → 6.7/1 ≈ 6.7 mol
O: 53.3 g → 53.3/16 ≈ 3.33 mol
Divide by smallest = 3.33 → C:1, H:2, O:1 → CH₂O
9. How many molecules are in 0.5 mol of H₂O?
A. 3.01 × 10²³
B. 6.02 × 10²³
C. 1.2 × 10²⁴
D. 0.5 × 10²³
Answer: A
Rationale: 0.5 × 6.022 ×10²³ = 3.01 ×10²³
Exam – Portage Learning – Academic
Year | Exam questions with verified
answers|| LATEST UPDATE 2026
CHEM 104 — MODULE 3 PRACTICE
QUESTIONS 2026
SECTION 1 — STOICHIOMETRY AND CHEMICAL
REACTIONS
1. How many moles are in 12 g of carbon?
A. 0.5 mol
B. 1 mol
C. 2 mol
D. 12 mol
Answer: B
Rationale: Molar mass of carbon = 12 g/mol → 12 g ÷ 12 g/mol = 1 mol
2. How many grams are in 2 moles of H₂O?
A. 36 g
B. 18 g
C. 9 g
D. 2 g
Answer: A
Rationale: Molar mass H₂O = 18 g/mol → 2 × 18 = 36 g
,3. What is the mass percent of O in H₂O?
A. 88.9%
B. 11.1%
C. 50%
D. 33%
Answer: A
Rationale: O mass = 16 g, H = 2 g → total = 18 g → %O = 16/18 × 100 ≈ 88.9%
4. Which is a balanced equation?
A. 2H₂ + O₂ → 2H₂O
B. H₂ + O₂ → H₂O
C. 2H₂ + O₂ → H₂O
D. H₂ + O → H₂O
Answer: A
Rationale: Both sides have 4 H atoms and 2 O atoms
5. How many moles of O₂ are needed to react with 4 moles of H₂ in 2H₂ + O₂ →
2H₂O?
A. 1 mole
B. 2 moles
C. 4 moles
D. 8 moles
Answer: A
Rationale: Ratio H₂:O₂ = 2:1 → 4 mol H₂ × (1/2) = 2 mol O₂ Wait—careful
Step: 2 H₂ + 1 O₂ → 2 H₂O
4 mol H₂ × (1 mol O₂ / 2 mol H₂) = 2 mol O₂ ✅
Correct Answer: B
6. In 2Na + Cl₂ → 2NaCl, the limiting reagent if 3 moles Na and 2 moles Cl₂ react
is:
A. Na
B. Cl₂
C. NaCl
, D. Both are in excess
Answer: B
Rationale: Ratio 2:1 → 3 mol Na needs 1.5 mol Cl₂ → 2 mol Cl₂ available → Cl₂ is in excess,
Na is limiting? Let's check:
2Na + Cl₂ → 2NaCl
3 mol Na × (1 mol Cl₂ / 2 mol Na) = 1.5 mol Cl₂ needed → 2 mol Cl₂ available → Na limits
reaction
✅ Limiting reagent = Na
7. How many grams of NaCl are produced from 1 mole Na + 1 mole Cl₂?
A. 58.44 g
B. 29.22 g
C. 117 g
D. 22.99 g
Answer: A
Rationale: 2Na + Cl₂ → 2NaCl → 1 mol Na reacts with 0.5 mol Cl₂ → forms 1 mol NaCl →
58.44 g
8. What is the empirical formula of a compound with 40% C, 6.7% H, 53.3% O?
A. CH₂O
B. C₂H₄O
C. CHO
D. C₂H₆O₂
Answer: A
Rationale:
C: 40 g → 40/12 ≈ 3.33 mol
H: 6.7 g → 6.7/1 ≈ 6.7 mol
O: 53.3 g → 53.3/16 ≈ 3.33 mol
Divide by smallest = 3.33 → C:1, H:2, O:1 → CH₂O
9. How many molecules are in 0.5 mol of H₂O?
A. 3.01 × 10²³
B. 6.02 × 10²³
C. 1.2 × 10²⁴
D. 0.5 × 10²³
Answer: A
Rationale: 0.5 × 6.022 ×10²³ = 3.01 ×10²³
