Question

The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below:
2KMno₄ + 10FeSO₄ + 8H₂O → K₂SO₄ + 2MnSO₄ + 5Fe2(SO₄)₃ + 8H₂O
If 15.8g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.

Answer 1

Molecular weight of KMnO₄ = 39 + 55 + 16 x 4 = 158
Molecular weight of K₂SO₄ = 2 x 39 + 32 + 16 x 4 = 174
Molecular weight of FeSO₄ = 56 + 32 + 64 = 152
2 x 158g of KMnO₄  yields= 174g of K₂SO₄.
So, 15.8g of KMnO₄ will yield = 174 x 15.8/2 x 158 = 8.7g of K₂SO₄ .
174g of K₂SO₄ yields 152g of FeSO₄
So, 8.7 g of K₂SO₄ will yield = 152 x 8.7/174 = 7.6 g of FeSO_4.
Hence, 7.6g of iron (II) sulphate is used in the above reaction.