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

41.1%

Step-by-step 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

`\boxed{\text{41.1 \%}}`

Step-by-step 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${_3}$} * \frac{\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${_3}$} * \frac{\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} * \frac{\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} = \frac{\text{ actual yield}}{\text{ theoretical yield}} * 100 \% = \frac{\text{3.65 g}}{\text{8.88 g}} * 100 \% = \textbf{41.1 \%}\\\\\text{The percent yield is }\boxed{\textbf{41.1 \%}}`