Question

The reaction of potassium permanganate with acidified iron (II) sulphate is given below :

2KMnO4 + 10FeSO4 + 8H2SO4 ⟶ K2SO4 + 2MnSO4 + 5 Fe2(SO4)3 + 8H2O.

If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.

[K = 39, Mn = 55, Fe = 56, S = 32, O = 16]

  • Percentage Composition
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Answer

2KMnO4 + 10FeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O

316 g of potassium permanganate was used with 1520 g of iron (II) sulphate

∴ 15.8 g. of potassium permanganate will be used with 1520\over316 × 15.8

= 76 g of iron (II) sulphate

Hence, 76 g of iron (II) sulphate was used.

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