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Q.

In the reaction ${CrO}_{5} + H_{2} {SO}_{4} \to {Cr}_{2} {({SO}_{4})}_{3} + H_{2} O + O_{2}$ one mole of ${CrO}_{5}$ will liberate how many moles of $O_{2}$

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a

9/2

b

5/2

c

7/2

d

5/4

answer is D.

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Detailed Solution

$2 {CrO}_{5} + 3 H_{2} {SO}_{4} \to {Cr}_{2} {({SO}_{4})}_{3} + 3 H_{2} O + 7 / 2 O_{2}$

1 mole ${CrO}_{5}$ liberates $\overset{}{\to}$ 7/2 mole of $O_{2}$

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