the chemical equation shows iron(iii) phosphate reacting with sodium sulfate. 2fepo4 + 3na2so4 → fe2(so4)3 +…

the chemical equation shows iron(iii) phosphate reacting with sodium sulfate. 2fepo4 + 3na2so4 → fe2(so4)3 + 2na3po4 what is the theoretical yield of fe2(so4)3 if 20.00 g of fepo4 reacts with an excess of na2so4? 26.52 g 53.04 g 150.8 g 399.9 g

the chemical equation shows iron(iii) phosphate reacting with sodium sulfate. 2fepo4 + 3na2so4 → fe2(so4)3 + 2na3po4 what is the theoretical yield of fe2(so4)3 if 20.00 g of fepo4 reacts with an excess of na2so4? 26.52 g 53.04 g 150.8 g 399.9 g

Answer

Explanation:

Step1: Calculate molar mass of FePO₄

The molar mass of Fe (55.85 g/mol), P (30.97 g/mol), O (16.00 g/mol). So molar mass of FePO₄ = 55.85+30.97 + 4×16.00=150.82 g/mol.

Step2: Calculate moles of FePO₄

Given mass of FePO₄ is 20.00 g. Moles of FePO₄ = $\frac{20.00\ g}{150.82\ g/mol}$≈0.1326 mol.

Step3: Determine mole - ratio

From the balanced equation 2FePO₄ + 3Na₂SO₄→Fe₂(SO₄)₃ + 2Na₃PO₄, the mole - ratio of FePO₄ to Fe₂(SO₄)₃ is 2:1. So moles of Fe₂(SO₄)₃ formed = $\frac{0.1326\ mol}{2}$ = 0.0663 mol.

Step4: Calculate molar mass of Fe₂(SO₄)₃

Molar mass of Fe₂(SO₄)₃ = 2×55.85+3×(32.07 + 4×16.00)=399.88 g/mol.

Step5: Calculate theoretical yield of Fe₂(SO₄)₃

Theoretical yield = moles×molar mass = 0.0663 mol×399.88 g/mol≈26.52 g.

Answer:

26.52 g