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The molar ratio of ${Fe}^{+ +} to {Fe}^{+ + +}$ in a mixture of ${FeSO}_{4}$ and ${Fe}_{2} {({SO}_{4})}_{3}$ having equal number of sulphate ions in both ferrous and ferric sulphate is

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1 : 2

3 : 2

2 : 3

can't be detrmined

answer is B.

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Detailed Solution

${FeSO}_{4}$

1 mole of ${SO}_{4}^{2 -} = 1 mole {Fe}^{2 +}$

In ${Fe}_{2} {({SO}_{4})}_{3}$

3 moles of ${SO}_{4}^{2 -} = 2 mole {Fe}^{3 +}$

1 mole of ${SO}_{4}^{2 -} = \frac{2}{3} moles {Fe}^{3 +}$

$ratio = \frac{{Fe}^{2 +}}{{Fe}^{3 +}} = \frac{1}{2} = \frac{3}{2}$

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