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Assume 0.10 L of $N_{2}$ and 0.18 L of $H_{2}$ , both at 50atm and 450°C, are reacted to form ${NH}_{3}$ . Assuming the reaction goes to completion, identify the reagent that is in excess and determine the volume of it that remains at the same temperature and pressure.

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$N_{2}, 0.04 L$

$H_{2}, 0 .08 L$

$H_{2}, 0.02 L$

$N_{2}, 0.01 L$

answer is D.

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Detailed Solution

In the reaction we have,

$\underset{0.10}{N_{2}} + \underset{0.18}{3 H_{2}} ⟶ 2 {NH}_{3}$

$\underset{= 0.04 L}{0.1 e - 0.06} - \underset{= 0.12 L}{2 \times 0.06}$

So the correct options is D.

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