Given the reaction: 2 FePO4 + 3 Na2SO4 → Fe2(SO4)3 + 2 Na3PO4, what is the theoretical yield (in grams) of Fe2(SO4)3 if 20.00 g of FePO4 reacts with excess Na2SO4?

• A. 12.0 g
• B. 36.2 g
• C. 24.5 g
• D. 18.1 g

Answer: (D)

Verified step by step guidance

1. Identify the given information: 20.00 g of FePO4 reacts with excess Na2SO4, and we want to find the theoretical yield of Fe2(SO4)3 in grams.
2. Calculate the molar mass of FePO4 by summing the atomic masses of Fe, P, and 4 O atoms. Use this molar mass to convert 20.00 g of FePO4 to moles using the formula: \(\text{moles FePO4} = \frac{\text{mass FePO4}}{\text{molar mass FePO4}}\).
3. Use the stoichiometric coefficients from the balanced chemical equation \$2 \text{FePO}_4 + 3 \text{Na}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 2 \text{Na}_3\text{PO}_4\( to find moles of Fe2(SO4)3 produced from moles of FePO4. The mole ratio is \)\frac{1 \text{ mole Fe}_2(\text{SO}_4)_3}{2 \text{ moles FePO}_4}$.
4. Calculate the molar mass of Fe2(SO4)3 by summing the atomic masses of 2 Fe atoms, 3 S atoms, and 12 O atoms. Then convert moles of Fe2(SO4)3 to grams using: \(\text{mass Fe}_2(\text{SO}_4)_3 = \text{moles Fe}_2(\text{SO}_4)_3 \times \text{molar mass Fe}_2(\text{SO}_4)_3\).
5. The result from step 4 is the theoretical yield of Fe2(SO4)3 in grams, assuming FePO4 is the limiting reagent and Na2SO4 is in excess.