Mole/Mass Problems - Free Printable

1. Balanced equation: 2Al + 3Cl₂ → 2AlCl₃

a. 2 moles of Al will react with 3 mole(s) of Cl₂ to produce 2 mole(s) of AlCl₃.

b. Moles of AlCl₃ produced from 2.50 moles Al:
From the balanced equation, 2 mol Al → 2 mol AlCl₃, so 2.50 mol Al → 2.50 mol AlCl₃.
Molar mass AlCl₃ = 27.0 + 3(35.5) = 133.5 g/mol.
Mass = 2.50 mol × 133.5 g/mol = 333.75 g.

c. Moles of Cl₂ needed to produce 12.3 g AlCl₃:
Moles AlCl₃ = 12.3 g / 133.5 g/mol ≈ 0.0921 mol.
From balanced equation, 2 mol AlCl₃ require 3 mol Cl₂.
Moles Cl₂ = (3/2) × 0.0921 mol ≈ 0.138 mol.

d. Grams of aluminum reacting with 3.4 moles Cl₂:
From balanced equation, 3 mol Cl₂ react with 2 mol Al.
Moles Al = (2/3) × 3.4 mol ≈ 2.267 mol.
Mass Al = 2.267 mol × 27.0 g/mol ≈ 61.2 g.

e. Moles of AlCl₃ produced from 17 g Al:
Moles Al = 17 g / 27.0 g/mol ≈ 0.6296 mol.
From balanced equation, 2 mol Al → 2 mol AlCl₃, so moles AlCl₃ = 0.6296 mol ≈ 0.63 mol.