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Q.

In the reaction $N_{2} H_{4} + 3 O_{2} \to 2 {NO}_{2} + 2 H_{2} O$ the mass of $O_{2}$ required to combine with $745 g$ of $N_{2} H_{4}$ will be

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a

$2510 g$

b

$2120 g$

c

$2436 g$

d

$2235 g$

answer is B.

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Detailed Solution

We have $\begin{aligned} N_{2} H_{4} + 3 O_{2} \to 2 {NO}_{2} + 2 H_{2} O \\ 32 g 96 g \end{aligned}$

Mass of $O_{2}$ required $= (\frac{96 g}{32 g}) (745 g) = 2235 g$

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