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Atomic Mass

The molar rat...

Question

The molar ratio of $F e^{2 +}$ to $F e^{3 +}$ in a mixture of $F e S O_{4}$ and $F e_{2} (S O_{4})_{3}$ having equal number of sulphate ions in both ferrous and ferric sulphates is:

1 : 2

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3 :2

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2 :3

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none of these

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Solution

The correct option is D 3 :2
1 mole of $F e S O_{4}$ and $F e_{2} (S O_{4})_{3}$ contain 1 and 3 moles respectively of sulphate ions.

For equal no. of sulphate ions, 3 moles of $F e S O_{4}$ are equivalent to 1 mole of $F e_{2} (S O_{4})_{3}$

Therefore required mole ratio $= \frac{3 \times 1}{1 \times 2} = \frac{3}{2}$

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Similar questions

F e^{2 +}

F e^{3 +}

F e S O_{4}

F e_{2} (S O_{4})_{3}

m o l e s

p e r m o l e

m o l e s

K M n O_{4}

p e r m o l e

K M n O_{4}

K M n O_{4} + H_{2} S O_{4} + F e S O_{4} \to K_{2} S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

5.5 g

F e S O_{4} \cdot 7 H_{2} O

F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O

5.4 m L

0.1 N K M n O_{4}

H = 1, O = 16, S = 32, F e = 56

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