Question

I would really appreciate it if someone helped me out with this question.

Answer 1

a) 0.085 moles of Fe₂(SO₄)₃ are produced.

b) Percent yield = 54.4%

Step-by-step 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%