Answer:
a) 0.085 moles of Fe₂(SO₄)₃ are produced.
b) Percent yield = 54.4%
Explanation:
Given data:
Moles of FePO₄ = 0.17mol
Number of moles of Fe₂(SO₄)₃ = ?
Solution:
Chemical equation:
2FePO₄ + 3Na₂SO₄ → Fe₂(SO₄)₃ + 2Na₃PO₄
Now we will compare the moles of FePO₄ with Fe₂(SO₄)₃
FePO₄ : Fe₂(SO₄)₃
2 : 1
0.17 : 1/2×0.17 = 0.085 mol
From 0.17 moles of FePO₄ 0.085 moles of Fe₂(SO₄)₃ are produced.
b)
Given data:
Mass of Fe₂(SO₄)₃ produced = 18.5 g
Percent yield = ?
Solution:
Theoretical yield of Fe₂(SO₄)₃:
Mass = number of moles × molar mass
Mass = 0.085 mol × 399.88 g/mol
Mass = 34 g
Percent yield:
Percent yield = (actual yield / theoretical yield )× 100
Percent yield = (18.5 g / 34 g) × 100
Percent yield = 54.4%
