The number of moles of $C r_{2} O_{7}^{2 -}$ needed to oxidize $0.136$ equivalents of $N_{2} H_{5}^{+}$ by the reaction is:
$N_{2} H_{5}^{+} + C r_{2} O_{7}^{2 -} \to N_{2} + C r^{3 +} + H_{2} O$

0.136

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0.068

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0.0227

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0.272

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Solution

The correct option is A $0.136$
$N_{2} H_{3}^{+} + C r_{2} O_{7}^{2 -} \to N_{2} C r^{+ 3}$
Balance the above reaction oxidation state of $N = - 2$
Increasing should be equal to decreasing
$3 N_{2} H_{5}^{+} + 2 C r_{2} O_{7}^{2 -} + 13 H^{+} \to N_{2} + 4 C r^{+ 3} + 14 H_{2} O$
1 mol of $N_{2} H_{5} = 2 \times$ No. equivalency
$2 \times 0.136 = 0.272$
$3 m o l N_{2} H_{5}^{+}$ required $2 m o l C r_{2} O_{7}^{2 -}$
$0.272 m o l N_{2} H_{5}^{+}$ required $\frac{2}{3} \times 0.272$
$= \frac{0.544}{3} = 0.181$

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