Chemistry worksheet focusing on limiting reactant problems, featuring four balanced chemical equations and questions about mass calculations, limiting reactants, and excess reactant amounts.

1. Balanced reaction: 2Al + 6HBr → 2AlBr₃ + 3H₂
a. Moles of Al = 86.9 g / 26.98 g/mol = 3.221 mol
Moles of HBr = 401 g / 80.91 g/mol = 4.955 mol
From the reaction, 2 mol Al requires 6 mol HBr.
Required HBr for 3.221 mol Al = (6/2) × 3.221 = 9.663 mol
But only 4.955 mol HBr available → HBr is limiting
Moles of H₂ formed = (3/6) × 4.955 = 2.4775 mol
Mass of H₂ = 2.4775 mol × 2.016 g/mol = 4.996 g ≈ 5.00 g
b. HBr is the limiting reactant
c. Moles of Al used = (2/6) × 4.955 = 1.6517 mol
Mass of Al used = 1.6517 mol × 26.98 g/mol = 44.57 g
Mass of Al left = 86.9 g - 44.57 g = 42.33 g

2. Balanced reaction: 3Si + 2N₂ → Si₃N₄
a. Moles of Si = 602 g / 28.09 g/mol = 21.43 mol
Moles of N₂ = 494 g / 28.02 g/mol = 17.63 mol
From the reaction, 3 mol Si requires 2 mol N₂.
Required N₂ for 21.43 mol Si = (2/3) × 21.43 = 14.29 mol
Available N₂ = 17.63 mol > 14.29 mol → Si is limiting
Moles of Si₃N₄ formed = (1/3) × 21.43 = 7.143 mol
Mass of Si₃N₄ = 7.143 mol × 140.39 g/mol = 1002.8 g ≈ 1003 g
b. Si is the limiting reactant
c. Moles of N₂ used = (2/3) × 21.43 = 14.29 mol
Mass of N₂ used = 14.29 mol × 28.02 g/mol = 400.5 g
Mass of N₂ left = 494 g - 400.5 g = 93.5 g

3. Balanced reaction: CuCl₂ + 2KI → CuI + 2KCl + I₂
a. Moles of CuCl₂ = 75.3 g / 134.45 g/mol = 0.5599 mol
Moles of KI = 106 g / 166.00 g/mol = 0.6386 mol
From the reaction, 1 mol CuCl₂ requires 2 mol KI.
Required KI for 0.5599 mol CuCl₂ = 2 × 0.5599 = 1.1198 mol
Available KI = 0.6386 mol < 1.1198 mol → KI is limiting
Moles of I₂ formed = (1/2) × 0.6386 = 0.3193 mol
Mass of I₂ = 0.3193 mol × 253.81 g/mol = 81.04 g
b. KI is the limiting reactant
c. Moles of CuCl₂ used = (1/2) × 0.6386 = 0.3193 mol
Mass of CuCl₂ used = 0.3193 mol × 134.45 g/mol = 42.92 g
Mass of CuCl₂ left = 75.3 g - 42.92 g = 32.38 g

4. Balanced reaction: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
a. Moles of FeS₂ = 1.35 g / 119.98 g/mol = 0.01125 mol
Moles of O₂ = 5.44 g / 32.00 g/mol = 0.1700 mol
From the reaction, 4 mol FeS₂ requires 11 mol O₂.
Required O₂ for 0.01125 mol FeS₂ = (11/4) × 0.01125 = 0.03094 mol
Available O₂ = 0.1700 mol > 0.03094 mol → FeS₂ is limiting
Moles of SO₂ formed = (8/4) × 0.01125 = 0.0225 mol
Mass of SO₂ = 0.0225 mol × 64.07 g/mol = 1.4416 g ≈ 1.44 g
b. FeS₂ is the limiting reactant
c. Moles of O₂ used = (11/4) × 0.01125 = 0.03094 mol
Mass of O₂ used = 0.03094 mol × 32.00 g/mol = 0.9901 g
Mass of O₂ left = 5.44 g - 0.9901 g = 4.45 g