Answer:
[tex]\boxed{\text{66.95 g BaSO$_{4}$}}[/tex]
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 261.34 233.39
Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃
m/g: 75.00
1. Moles of Ba(NO₃)₂
[tex]\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}[/tex]
2. Moles of BaSO₄
The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂
[tex]\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}[/tex]
3. Mass of BaSO₄
[tex]\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}[/tex]
