Question

The molar ratio of $F{e}^{2+}$ to $F{e}^{3+}$ in a mixture of $FeS{O}_4$ and $F{e}_2(S{O}_4{)}_3$ having equal number of sulphate ion in both ferrous and ferric sulphate is :

• A. 1 : 2
• B. 3 : 2
• C. 3:1
• D. can't be determined

Answer 1

The correct option is C 3:1

Suppose we start with 1 mole of ferric sulphate $F{e}_2(S{O}_4{)}_3$. It will contain 3 moles of sulphate ions.

Ferrous sulphate $FeS{O}_4$ will also have 3 moles of sulphate ions.

1 mole of ferrous sulphate contains 1 mole of sulphate ions. 3 moles of sulphate ions will be present in 3 moles of ferrous sulphate.

Hence, the molar ratio is ferrous sulphate : ferric sulphate $=3:1$

Note: Number of sulphate ions in both ferrous and ferric sulphate is equal. Hence, number of moles of sulphate ions in both ferrous and ferric sulphate is equal.