Question

For the following reaction, 4.53 grams of calcium hydroxide are mixed with excess sulfuric acid. The reaction yields 6.75 grams of calcium sulfate.
sulfuric acid (aq) + calcium hydroxide (s) calcium sulfate (s) + water (l)

What is the theoretical yield of calcium sulfate? (grams)
What is the percent yield for this reaction? (%)

Answer 1

Answer:

$\large \boxed{\text{8.32 g; 81.1 \%}}}$

Explanation:

MM:                      74.09         136.14

           H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O

m/g:                      4.53  

1. Theoretical yield

(a) Moles of Ca(OH)₂

$\text{Moles of Ca(OH) {_2} } = \text{4.53 g Ca(OH) {_2} } \times \dfrac{\text{1 mol Ca(OH) {_2} }}{\text{74.09 g Ca(OH) {_2} }} = \text{0.161 14 mol Ca(OH) {_2} }$

(b) Moles of CaSO₄

$\text{Moles of NH {_3} } = \text{0.061 14 mol Ca(OH) {_2} } \times \dfrac{\text{1 mol CaSO {_4} }}{\text{1 mol Ca(OH) {_2} }} = \text{0.061 14 mol CaSO {_4} }$

(c) Theoretical yield

$\text{Mass of CaSO {_4} } = \text{0.061 14 mol CaSO {_4} } \times \dfrac{\text{136.14 g CaSO {_4} }}{\text{1 mol CaSO {_4} }} = \textbf{8.32 g CaSO {_4} }$

2. Percent yield

$\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \% = \dfrac{\text{6.75 g}}{\text{8.32 g}} \times 100 \, \% = \text{81.1 \%}$

$\text{The percent yield is } \large \boxed{\textbf{81.1 \%}}}$