# Question 19e89

Jan 16, 2014

You divide the actual yield by the theoretical yield and multiply by 100.

EXAMPLE

What is the percent yield if 0.650 g of copper are formed when excess aluminum reacts with a 2.00 g of copper(II) chloride dihydrate according to the equation

Step 1. Write the balanced chemical equation

3CuCl₂·2H₂O + 2Al → 3Cu + 2AlCl₃ + 2H₂O

Step 2. Calculate the theoretical yield of Cu

(a) Calculate the moles of 3CuCl₂·2H₂O

$\text{2.00 g CuCl"_2·2"H"_2"O" × ("1 mol CuCl"_2·2"H"_2"O")/("170.5 g CuCl"_2·2"H"_2"O") = "0.011 73 mol CuCl"_2·2"H"_2"O}$

(b) Calculate the moles of Cu

$\text{0.011 73 mol CuCl"_2·2"H"_2"O" × "3 mol Cu"/("3 mol CuCl"_2·2"H"_2"O") = "0.011 73 mol Cu}$

(c) Calculate the theoretical yield

$\text{0.011 73 mol Cu"× "63.55 g Cu"/"1 mol Cu" = "0.7455 g Cu}$

Step 3. Calculate the percent yield.

% yield = "theoretical yield"/"actual yield" × 100 % = "0.650 g"/"0.7455 g" × 100 %# = 87.2 %