Question

How many moles of `NH_3` are produced when 1.2 moles `H_2` reacts?

Answer 1

`N_2(g) + 3H_2(g) rarr 2NH_3(g)`. Approx. `0.8` `mol` ammonia are produced.

Explanation:

We assume (very unreasonably!) quantitative reaction.

The stoichiometry of the reaction dictates that for each equiv dinitrogen, 3 equiv of dihydrogen react, to give 2 equiv ammonia.

Since there are `1.2` `mol` dihydrogen, `0.4` `mol` dinitrogen could have reacted; thus, an `0.8` `mol` quantity of ammonia could have been produced.

All I am doing is following the reaction stoichiometry. Note that such a quantitative reaction in dinitrogen reduction is very unfeasible, and a chemical engineer would jump for joy if such turnovers were achieved.