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

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%

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~batmans wife dun dun