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Chemistry: Stoichiometry Practice Problems

Worked mole, mass, limiting reactant, percent yield, empirical formula and solution problems. Every explanation shows the full calculation.

by Kill h.18 questionsCreated Nov 8, 2026, 08:03 PM

Chemistry· Stoichiometry#stoichiometry#moles#chemistry#practice-problems

18 questions

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1
What is the molar mass of water (H₂O)? Use H = 1.008 and O = 16.00 g/mol.
- A
  17.01 g/mol
- B
  10.00 g/mol
- 18.02 g/mol
- D
  34.02 g/mol
. 17.01 g/mol counts only one hydrogen (that is OH), and 34.02 g/mol is hydrogen peroxide, H₂O₂.
2
How many molecules are in 0.500 mol of CO₂?
- A
- C
- D
Molecules = moles × Avogadro's number = . would be a full mole.
3
How many oxygen atoms are in 1.00 mol of CO₂?
- A
- B
- C
Each CO₂ molecule has 2 O atoms, so 1.00 mol CO₂ contains 2.00 mol O atoms: . counts all three atoms.
4
What is the molar mass of calcium carbonate, CaCO₃? (Ca = 40.08, C = 12.01, O = 16.00)
- 100.09 g/mol
- B
  68.09 g/mol
- C
  56.08 g/mol
- D
  116.09 g/mol
. 68.09 g/mol uses only one oxygen, and 56.08 g/mol is CaO.
5
Balance the combustion of propane: . What is the coefficient of O₂?
- A
  3
- B
  4
- C
  7
- 5
Count O on the right: 3 CO₂ gives 6 O and 4 H₂O gives 4 O, total 10 O atoms, so 5 O₂. Balanced: .
6
For , what mass of water forms from 4.032 g of H₂ with excess oxygen? (H₂ = 2.016 g/mol, H₂O = 18.02 g/mol)
- A
  18.0 g
- 36.0 g
- C
  72.1 g
- D
  4.03 g
Moles H₂ = 4.032 / 2.016 = 2.000 mol. The mole ratio H₂ : H₂O is 2 : 2, so 2.000 mol H₂O forms. Mass = . Mass is not conserved molecule by molecule, so you cannot skip the mole step.
7
For , you start with 2.0 mol N₂ and 3.0 mol H₂. Which reactant is limiting?
- H₂
- B
  N₂
- C
  Neither; they are in the exact ratio
- D
  NH₃
2.0 mol N₂ would need 3 × 2.0 = 6.0 mol H₂, but only 3.0 mol is available, so H₂ runs out first. Having fewer moles is not the test: you must compare with the mole ratio.
8
Using the same reaction () with 2.0 mol N₂ and 3.0 mol H₂, how many moles of NH₃ can form?
- 2.0 mol
- B
  4.0 mol
- C
  3.0 mol
- D
  6.0 mol
Base the answer on the limiting reactant, H₂: . 4.0 mol is what you get if you wrongly start from N₂.
9
A reaction has a theoretical yield of 50.0 g, but you collect 42.0 g of product. What is the percent yield?
- A
  119%
- B
  16.0%
- 84.0%
- D
  8.0%
Percent yield = actual / theoretical × 100 = . Dividing the other way gives 119%, which is impossible without impurities or wet product.
10
A compound is 40.0% C, 6.7% H and 53.3% O by mass. What is its empirical formula?
- A
  CHO
- CH₂O
- C
  C₂H₄O
- D
  CH₃O₂
Assume 100 g: C 40.0 / 12.01 = 3.33 mol, H 6.7 / 1.008 = 6.65 mol, O 53.3 / 16.00 = 3.33 mol. Divide by the smallest (3.33): C 1, H 2, O 1, so CH₂O.
11
The compound with empirical formula CH₂O has a molar mass of 180 g/mol. What is its molecular formula?
- C₆H₁₂O₆
- B
  C₃H₆O₃
- C
  CH₂O
- D
  C₁₂H₂₂O₁₁
The mass of CH₂O is about 30.03 g/mol, and 180 / 30.03 ≈ 6. Multiply every subscript by 6 to get C₆H₁₂O₆ (glucose). C₃H₆O₃ is only 90 g/mol.
12
What is the percent by mass of oxygen in water?
- A
  33.3%
- B
  11.2%
- 88.8%
- D
  50.0%
% O = 16.00 / 18.02 × 100 = 88.8%. 33.3% counts atoms (1 of 3), not mass. Oxygen is much heavier than hydrogen, so it dominates the mass.
13
5.85 g of NaCl (58.44 g/mol) is dissolved to make 0.500 L of solution. What is the molarity?
- A
  0.100 M
- 0.200 M
- C
  11.7 M
- D
  0.0500 M
Moles NaCl = 5.85 / 58.44 = 0.100 mol. Molarity = 0.100 mol / 0.500 L = 0.200 M. 0.100 M forgets to divide by the volume.
14
50.0 mL of 6.0 M HCl is diluted with water to a final volume of 300. mL. What is the new concentration?
- A
  36 M
- 1.0 M
- C
  0.17 M
- D
  1.2 M
: , so . 1.2 M is what you get if you wrongly use 250 mL of added water as the final volume.
15
What volume does 0.250 mol of an ideal gas occupy at STP (0 °C and 1 atm)?
- A
  22.4 L
- B
  89.6 L
- C
  0.0112 L
- 5.60 L
At 0 °C and 1 atm, one mole of an ideal gas occupies 22.4 L: . Note that the newer IUPAC STP (1 bar) uses 22.7 L/mol.
16
How many grams of CO₂ form when 16.04 g of methane burns completely? (CH₄ = 16.04, CO₂ = 44.01 g/mol)
- A
  16.0 g
- 44.0 g
- C
  88.0 g
- D
  22.0 g
Moles CH₄ = 16.04 / 16.04 = 1.000 mol. The ratio CH₄ : CO₂ is 1 : 1, so 1.000 mol CO₂ = 44.0 g. 88.0 g wrongly uses the 2 from the O₂ coefficient.
17
For , what mass of NaCl forms from 46.0 g of sodium with excess chlorine? (Na = 22.99, NaCl = 58.44 g/mol)
- A
  58.4 g
- B
  81.5 g
- C
  234 g
- 117 g
Moles Na = 46.0 / 22.99 = 2.00 mol. The ratio Na : NaCl is 2 : 2, so 2.00 mol NaCl × 58.44 = 117 g. 81.5 g just adds the masses of one Na and one Cl, which is not how stoichiometry works.
18
What is the molar mass of copper(II) sulfate pentahydrate, CuSO₄·5H₂O? (CuSO₄ = 159.61, H₂O = 18.02 g/mol)
- 249.7 g/mol
- B
  177.6 g/mol
- C
  159.6 g/mol
- D
  339.8 g/mol
Add five waters of hydration: . 177.6 g/mol includes only one water.