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Stoichiometric Calculations

How many gram...

Question

How many grams of $K_{2} C r_{2} O_{7}$ is required to oxidise $12.41$ g of $F e S O_{4}$ to $F e_{2} (S O_{4})_{3}$ in an acidic medium? $(K_{2} C r_{2} O_{7} = 294; F e S O_{4} = 152)$

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Solution

The reaction is:
$6 F e S O_{4} + K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} \to 3 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 7 H_{2} O$

12.41 g of $F e S O_{4}$ corresponds to 0.082 mole of $F e S O_{4}$ .

Mole of $K_{2} C r_{2} O_{7}$ required is $\frac{0.082}{6}$ .

Hence, grams of potassium dichromate = $\frac{0.082}{6} \times 294 = 4.018$ gms

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