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What are the molar ratios of reactants and products in dinitrogen reduction by dihydrogen?

1 Answer

May 1, 2017

Well, $dinitrogen : dihydrogen : ammonia = 1 : 3 : 2 .$ $"dinitrogen : dihydrogen : ammonia = 1 : 3 : 2".$

Explanation:

And thus the molar ratio follows the given stoichiometry of the reaction:

$N_{2} + 3 H_{2} (g) \to 2 N H_{3} (g)$ $N_2 + 3H_2(g) rarr 2NH_3(g)$

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