Question

a lab group completely reacted to a 0.5 f piece of aluminum in excess of CuCl2. instead of rinsing off the copper, the group weighed it and obtained 1.50 g Cu. Calculate the percent yield of copper in their reaction. write the balanced chemical equation.​

Answer 1

Answer: 2Al + 3CuCl2 → 2AlCl3 + 3Cu, Percent yield is 84.9%

Explanation: Find moles of aluminum using molar mass then x by coefficient ratio of cu/al. Then divide 1.5 by that value and x 100.

Answer 2

1. Percent yield = (Actual yield/Theoretical yield) x 100

Actual yield = 1.50 g Cu

Theoretical yield = (0.5 f Al) x (1 mol Al/27.0 g Al) x (1 mol Cu/1 mol Al) x (63.5 g Cu/1 mol Cu) = 1.35 g Cu

Percent yield = (1.50 g Cu/1.35 g Cu) x 100 = 111.1%

2. Balanced chemical equation: 2 Al + 3 CuCl2 → 2 AlCl3 + 3 Cu