What is the maximum mass (in grams) of NO that could be obtained from 15.5 g of $N_{2} O_{4}$ and 4.68 g of $N_{2} H_{4}$ when they react? The balanced chemical equation is :
$2 N_{2} O_{4} + N_{2} H_{4} ⟶ 6 NO + 2 H_{2} O$

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5.04

26.2

4.38

15.2

answer is C.

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Detailed Solution

$\begin{aligned} 2 N_{2} O_{4} + N_{2} H_{4} \to 6 NO + 2 H_{2} O \\ n = \frac{15.5}{92} \frac{4.68}{32} \\ (LR) \frac{15.5}{92} \times \frac{1}{2} \times 6 \times 30 \\ = 15.16 g \end{aligned}$

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