Question #19e89

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Question #19e89

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Answer 1 · 2014-01-16T00:26:53

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

${2.00 g CuCl}_{2} \cdot 2 H_{2} O \times \frac{{1 mol CuCl}_{2} \cdot 2 H_{2} O}{{170.5 g CuCl}_{2} \cdot 2 H_{2} O} = {0.011 73 mol CuCl}_{2} \cdot 2 H_{2} O$ $"2.00 g CuCl"_2·2"H"_2"O" × ("1 mol CuCl"_2·2"H"_2"O")/("170.5 g CuCl"_2·2"H"_2"O") = "0.011 73 mol CuCl"_2·2"H"_2"O"$

(b) Calculate the moles of Cu

${0.011 73 mol CuCl}_{2} \cdot 2 H_{2} O \times \frac{3 mol Cu}{{3 mol CuCl}_{2} \cdot 2 H_{2} O} = 0.011 73 mol Cu$ $"0.011 73 mol CuCl"_2·2"H"_2"O" × "3 mol Cu"/("3 mol CuCl"_2·2"H"_2"O") = "0.011 73 mol Cu"$

(c) Calculate the theoretical yield

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

Step 3. Calculate the percent yield.

% yield = $\frac{theoretical yield}{actual yield} \times 100 % = \frac{0.650 g}{0.7455 g} \times 100 %$ $"theoretical yield"/"actual yield" × 100 % = "0.650 g"/"0.7455 g" × 100 %$ = 87.2 %

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Question #19e89

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