Answer:
[tex]\boxed{\text{41.1 \%}}[/tex]
Explanation:
MM: 100.09 18.02
CaCO₃ + H₂SO₄ ⟶ CaSO₄ + H₂O + CO₂
m/g: 49.3 3.65
1. Theoretical yield
(a) Moles of CaCO₃
[tex]\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${_3}$}[/tex]
(b) Moles of H₂O
[tex]\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}[/tex]
(c) Theoretical mass of H₂O
[tex]\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}[/tex]
(d) Percent yield
[tex]\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 \%}}[/tex]
