Answer:
[tex]\large \boxed{\text{8.32 g; 81.1 \%}}}[/tex]
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
[tex]\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)${_2}$}[/tex]
(b) Moles of CaSO₄
[tex]\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${_4}$}[/tex]
(c) Theoretical yield
[tex]\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${_4}$}[/tex]
2. Percent yield
[tex]\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 \%}[/tex]
[tex]\text{The percent yield is } \large \boxed{\textbf{81.1 \%}}}[/tex]
