Question #710f4

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Question #710f4

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Answer 1 · 2017-01-11T16:44:43

$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (l)$ $4NH_3(g) + 5O_2(g) rarr 4NO(g) + 6H_2O(l)$

Explanation:

Nitrous oxide is commonly an oxidation product of ammonia. If this is the reaction you propose, then $3$ $3$ $equiv$ $"equiv"$ of water result from $2$ $2$ $equiv$ $"equiv"$ of ammonia.

You react $40 \cdot m o l$ $40*mol$ of $N H_{3}$ $NH_3$ , then clearly you get $60 \cdot m o l$ $60*mol$ of $O H_{2}$ $OH_2$ . I acknowledge that this might not have been the reaction you had in mind.

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Question #710f4

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