the chemical equation shows iron(iii) phosphate reacting with sodium sulfate.\n2fepo₄ + 3na₂so₄ → fe₂(so₄)₃…

the chemical equation shows iron(iii) phosphate reacting with sodium sulfate.\n2fepo₄ + 3na₂so₄ → fe₂(so₄)₃ + 2na₃po₄\nwhat is the theoretical yield of fe₂(so₄)₃ if 20.00 g of fepo₄ reacts with an excess of na₂so₄?\n26.52 g\n53.04 g\n150.8 g\n399.9 g

the chemical equation shows iron(iii) phosphate reacting with sodium sulfate.\n2fepo₄ + 3na₂so₄ → fe₂(so₄)₃ + 2na₃po₄\nwhat is the theoretical yield of fe₂(so₄)₃ if 20.00 g of fepo₄ reacts with an excess of na₂so₄?\n26.52 g\n53.04 g\n150.8 g\n399.9 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). 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, $n_{FePO_4}=\frac{20.00\ g}{150.82\ g/mol}\approx0.1326\ mol$.

Step3: Determine mole - ratio from balanced equation

From the balanced equation $2FePO_{4}+3Na_{2}SO_{4}\to Fe_{2}(SO_{4}){3}+2Na{3}PO_{4}$, the mole - ratio of $FePO_{4}$ to $Fe_{2}(SO_{4}){3}$ is 2:1. So moles of $Fe{2}(SO_{4}){3}$, $n{Fe_2(SO_4)_3}=\frac{0.1326\ mol}{2}= 0.0663\ mol$.

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. Theoretical yield of $Fe_{2}(SO_{4})_{3}=0.0663\ mol\times399.88\ g/mol\approx26.52\ g$.

Answer:

26.52 g