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

\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 $\frac{7}{4}$
Balanced reaction is
$4 C r O_{5} + 6 H_{2} S O_{4} \to 2 C r_{2} (S O_{4})_{3} + 6 H_{2} O + 7 O_{2}$
1 mole $C r O_{5}$ produces $\frac{7}{4}$ moles of $O_{2}$

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