To determine the theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex] when 20.00 grams of [tex]\( FePO_4 \)[/tex] reacts with an excess of [tex]\( Na_2SO_4 \)[/tex], we should follow these steps:
1. Calculate the molar mass of [tex]\( FePO_4 \)[/tex]:
The molar mass of [tex]\( FePO_4 \)[/tex] is given as 150.82 g/mol.
2. Find the number of moles of [tex]\( FePO_4 \)[/tex]:
[tex]\[
\text{Number of moles of } FePO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{20.00 \text{ g}}{150.82 \text{ g/mol}} \approx 0.1326 \text{ moles}
\][/tex]
3. Use the stoichiometry of the balanced equation:
According to the balanced chemical equation
[tex]\[
2 FePO_4 + 3 Na_2SO_4 \rightarrow Fe_2(SO_4)_3 + 2 Na_3PO_4
\][/tex]
2 moles of [tex]\( FePO_4 \)[/tex] produce 1 mole of [tex]\( Fe_2(SO_4)_3 \)[/tex].
4. Calculate the moles of [tex]\( Fe_2(SO_4)_3 \)[/tex] produced:
[tex]\[
\text{Moles of } Fe_2(SO_4)_3 = \frac{\text{moles of } FePO_4}{2} = \frac{0.1326 \text{ moles}}{2} \approx 0.0663 \text{ moles}
\][/tex]
5. Calculate the molar mass of [tex]\( Fe_2(SO_4)_3 \)[/tex]:
The molar mass of [tex]\( Fe_2(SO_4)_3 \)[/tex] is given as 399.88 g/mol.
6. Calculate the theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex]:
[tex]\[
\text{Theoretical yield} = \text{moles of } Fe_2(SO_4)_3 \times \text{molar mass of } Fe_2(SO_4)_3 = 0.0663 \text{ moles} \times 399.88 \text{ g/mol} \approx 26.51 \text{ g}
\][/tex]
Therefore, the theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex] when 20.00 g of [tex]\( FePO_4 \)[/tex] reacts with an excess of [tex]\( Na_2SO_4 \)[/tex] is approximately [tex]\( 26.51 \)[/tex] grams.
Among the given choices:
- 26.52 g (closest to the calculated theoretical yield)
- 53.04 g
- 150.8 g
- 399.9 g
The correct answer is: 26.52 g.
