CART - Chemistry Acceleration Readiness Stoichiometry and Mole Concepts Questions and Answers — Questions and Answers
Question 1: What is the molar mass of iron(III) sulfate, Fe₂(SO₄)₃?
- 207.75 g/mol
- 349.91 g/mol
- 399.88 g/mol (Correct answer)
- 407.87 g/mol
Correct answer: 399.88 g/mol
To calculate the molar mass of Fe₂(SO₄)₃, you sum the atomic masses of all atoms in the formula. Iron (Fe) has a molar mass of approximately 55.845 g/mol. Sulfur (S) is ~32.065 g/mol, and Oxygen (O) is ~15.999 g/mol. The calculation is: (2 × 55.845) + 3 × (32.065 + 4 × 15.999) = 111.69 + 3 × (96.061) = 111.69 + 288.183 = 399.873 g/mol, which rounds to 399.88 g/mol.
Question 2: In the combustion of methane (CH₄), how much carbon dioxide (CO₂) in grams is produced from the complete combustion of 32.0 grams of methane? The balanced equation is: CH₄ + 2O₂ → CO₂ + 2H₂O
- 44.0 g
- 176.0 g
- 22.0 g
- 88.0 g (Correct answer)
Correct answer: 88.0 g
First, convert the mass of CH₄ to moles: Molar mass of CH₄ is 12.01 + 4(1.01) = 16.05 g/mol. So, 32.0 g / 16.05 g/mol ≈ 2.0 moles of CH₄. According to the balanced equation, the mole ratio of CH₄ to CO₂ is 1:1. Therefore, 2.0 moles of CO₂ are produced. Finally, convert moles of CO₂ to grams: Molar mass of CO₂ is 12.01 + 2(16.00) = 44.01 g/mol. So, 2.0 moles × 44.01 g/mol = 88.02 g. The closest answer is 88.0 g.
Question 3: A student reacts 25.0 g of nitrogen (N₂) with 5.0 g of hydrogen (H₂) to produce ammonia (NH₃) according to the equation: N₂ + 3H₂ → 2NH₃. Which of the following statements is correct?
- Nitrogen (N₂) is the limiting reactant, and excess hydrogen (H₂) remains.
- Hydrogen (H₂) is the limiting reactant, and excess nitrogen (N₂) remains. (Correct answer)
- Both reactants are completely consumed.
- Ammonia (NH₃) is the limiting reactant.
Correct answer: Hydrogen (H₂) is the limiting reactant, and excess nitrogen (N₂) remains.
To find the limiting reactant, calculate the moles of each reactant and compare it to the stoichiometric ratio. Moles of N₂ = 25.0 g / 28.02 g/mol ≈ 0.892 mol. Moles of H₂ = 5.0 g / 2.02 g/mol ≈ 2.48 mol. The required mole ratio of N₂ to H₂ is 1:3. Moles of H₂ needed to react with all the N₂ is 0.892 mol N₂ × 3 = 2.676 mol H₂. Since we only have 2.48 mol of H₂, hydrogen is the limiting reactant and will be consumed first, leaving excess nitrogen.
Question 4: How many individual carbon atoms are present in a 24.02 gram sample of pure carbon?
- 6.022 x 10²³ atoms
- 3.011 x 10²³ atoms
- 1.204 x 10²⁴ atoms (Correct answer)
- 2.408 x 10²⁴ atoms
Correct answer: 1.204 x 10²⁴ atoms
First, determine the number of moles in the sample. The molar mass of carbon is approximately 12.01 g/mol. Moles = Mass / Molar Mass = 24.02 g / 12.01 g/mol = 2.0 moles. Next, use Avogadro's number (6.022 x 10²³ atoms/mol) to find the number of atoms. Number of atoms = Moles × Avogadro's number = 2.0 mol × (6.022 x 10²³ atoms/mol) = 1.2044 x 10²⁴ atoms.
Question 5: In a laboratory experiment, the reaction of 15.0 g of aluminum (Al) with excess hydrochloric acid (HCl) produced 1.55 g of hydrogen gas (H₂). What is the percent yield of hydrogen? The balanced equation is: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
- 92.3% (Correct answer)
- 10.3%
- 1.68%
- 88.6%
Correct answer: 92.3%
First, calculate the theoretical yield. Moles of Al = 15.0 g / 26.98 g/mol ≈ 0.556 mol. From the 2:3 mole ratio between Al and H₂, moles of H₂ = 0.556 mol Al × (3 mol H₂ / 2 mol Al) = 0.834 mol H₂. Mass of H₂ (theoretical) = 0.834 mol × 2.02 g/mol = 1.68 g. The percent yield is (Actual Yield / Theoretical Yield) × 100 = (1.55 g / 1.68 g) × 100 ≈ 92.3%.
Question 6: Which of the following samples contains the greatest number of moles of the substance?
- 60.0 g of C₂H₆ (Molar Mass ≈ 30 g/mol)
- 90.0 g of H₂O (Molar Mass ≈ 18 g/mol) (Correct answer)
- 100.0 g of O₂ (Molar Mass ≈ 32 g/mol)
- 200.0 g of CaCO₃ (Molar Mass ≈ 100 g/mol)
Correct answer: 90.0 g of H₂O (Molar Mass ≈ 18 g/mol)
To find the number of moles, divide the mass of the sample by its molar mass. A) 60.0 g / 30 g/mol = 2.0 mol. B) 90.0 g / 18 g/mol = 5.0 mol. C) 100.0 g / 32 g/mol = 3.125 mol. D) 200.0 g / 100 g/mol = 2.0 mol. Comparing the results, 90.0 g of H₂O contains the greatest number of moles (5.0 mol).
CART - Chemistry Acceleration Readiness Stoichiometry and Mole Concepts Questions and Answers
What is the molar mass of iron(III) sulfate, Fe₂(SO₄)₃?