Question

A 49.3 sample of CaCO3 was treated with aqueous H2SO4, producing calcium sulfate, 3.65 g of water and CO2(g). What was the % yield of H2O?

Answer 1

Answer:

41.1%

Explanation:

First write the balanced reaction:

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Now calculate the theoretical yield:

49.3 g CaCO₃ × (1 mol CaCO₃ / 100 g CaCO₃) = 0.493 mol CaCO₃

0.493 mol CaCO₃ × (1 mol H₂O / 1 mol CaCO₃) = 0.493 mol H₂O

0.493 mol H₂O × (18 g H₂O / mol H₂O) = 8.87 g H₂O

Now calculate the % yield:

3.65 g H₂O / 8.87 g H₂O × 100% = 41.1%

Answer 2

Answer:

$\boxed{\text{41.1 \%}}$

Explanation:

MM: 100.09                                   18.02

       CaCO₃ + H₂SO₄ ⟶ CaSO₄ + H₂O + CO₂

m/g:  49.3                                        3.65  

1. Theoretical yield

(a) Moles of CaCO₃

$\text{Moles of CaCO {_3} } = \text{49.3 g CaCO {_3} } \times \dfrac{\text{1 mol CaCO {_3} }}{\text{100.09 g CaCO {_3} }} = \text{0.4926 mol CaCO {_3} }$

(b) Moles of H₂O

$\text{Moles of H {_2} O} = \text{0.4926 mol CaCO {_3} } \times \dfrac{\text{1 mol H {_2} O}}{\text{1 mol CaCO {_3} }} = \text{0.4926 mol H {_2} O}$

(c) Theoretical mass of H₂O

$\text{Mass of H {_2} O} = \text{0.4926 mol H {_2} O} \times \dfrac{\text{18.02 g H _{2} O}}{\text{1 mol H {_2} O}} = \text{8.88 g H {_2} O}$

(d) Percent yield

$\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \% = \dfrac{\text{3.65 g}}{\text{8.88 g}} \times 100 \% = \textbf{41.1 \%}\\\\\text{The percent yield is }\boxed{\textbf{41.1 \%}}$