Answer to Question #346527 in General Chemistry for Jose

Question

2 Al + 3 CuCl₂ → 2 AICI3 + 3 Cu

Molar masses: 27.0 g/mol Al,

134.5 g/mol CuCl2, 63.5 g/mol Cu

A: How many grams of CuCl₂ are

necessary to consume 10 g Al?

B: How many grams of Cu will be

produced?

Answer 1

Solution:

The molar mass of Al is 27.0 g/mol

Therefore,

Moles of Al = (10 g Al) × (1 mol Al / 27.0 g Al) = 0.3704 mol Al

Balanced chemical equation:

2Al + 3CuCl₂ → 2AICI₃ + 3Cu

According to stoichiometry:

2 mol of Al react with 3 mol of CuCl₂

Thus, 0.3704 mol of Al react with:

(0.3704 mol Al) × (3 mol CuCl₂ / 2 mol Al) = 0.5556 mol CuCl₂

The molar mass of CuCl₂ is 134.5 g/mol

Therefore,

Mass of CuCl₂ = (0.5556 mol CuCl₂) × (134.5 g CuCl₂ / 1 mol CuCl₂) = 74.7282 g CuCl₂ = 74.73 g CuCl₂

Thus, 74.73 grams of CuCl₂ are necessary to consume 10 g Al

According to stoichiometry:

2 mol of Al produce 3 mol of Cu

Thus, 0.370 mol of Al produce:

(0.3704 mol Al) × (3 mol Cu / 2 mol Al) = 0.5556 mol Cu

The molar mass of Cu is 63.5 g/mol

Therefore,

Mass of Cu = (0.5556 mol Cu) × (63.5 g Cu / 1 mol Cu) = 35.2806 g Cu = 35.28 g Cu

Thus, 35.28 grams of Cu will be produced

Answer:

A: 74.73 grams of CuCl₂

B: 35.28 grams of Cu