In the reaction given below, one mole of $C r O_{5}$ will liberate how many moles of oxygen?
$C r O_{5} + H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + H_{2} O + O_{2}$

\frac{5}{2}

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\frac{5}{4}

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\frac{9}{2}

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\frac{7}{4}

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Solution

The correct option is D $\frac{7}{4}$
$2 C r O_{5} + 3 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 3 H_{2} O + \frac{7}{2} O_{2}$
$1$ mole of $C r O_{5}$ gives $0.5$ moles of $C r_{2} (S O_{4})_{3}$ for which it requires $1.5$ moles of $H_{2} S O_{4}$ .
Moles of $O_{2}$ formed $= \frac{7}{4}$ .

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