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28g of $N_{2}$ is mixed with 28g of $H_{2}$ and they are made to react with another. Then the maximum weight of Ammonia obtained is

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$56 g$

$24 g$

$34 g$

$44 g$

answer is B.

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Detailed Solution

$\begin{array}{l} N_{2} + 3 H_{2} \to 2 N H_{3} \\ 28 g 6 g 34 g \\ 28 g 28 g ? \end{array}$

$N_{2}$ is limiting reagent
$\begin{array}{l} 28 g o f N_{2} = 34 g o f N H_{3} \\ 28 g o f N_{2} = ? \end{array}$
Amount of $N H_{3}$ = $\frac{34 \times 28}{28} = 34 g$ formed

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