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Large quantities of ammonia are burned in the presence of a platinum catalyst to give nitric oxide, as the first step in the preparation of nitric acid, ${NH}_{3} (g) + O_{2} (g) ⟶ NO (g) + H_{2} O (g)$ (unbalanced) Suppose a vessel contains 0.12 mo) ${NH}_{3}$ and 0.14 mol $O_{2}$ . How many moles of NO may be obtained?

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0.120

0.112

0.140

0.070

answer is B.

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Detailed Solution

In the given reaction we have;

$2 {NH}_{3} + \frac{5}{2} O_{2} \to 2 LR + 3 H_{2} O$

$n = 0.12 0.14 - -$

$(LR) (\frac{0.14}{5 / 2}) \times 2 = 0.112 mole$

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