How many moles of $N_2$ reacted if 0.75 mol of $NH_3$ is produced?

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Answer 1 · 2016-11-03T10:13:28

$1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)$ ; $0.38*mol$ $N_2$ reacted.

Given the stoichiometrically balanced equation, if $0.75*mol$ ammonia were produced, then at least $(0.75*mol)/2$ dinitrogen reacted. Is this clear?

How many moles of N_2N_2 reacted if 0.75 mol of NH_3NH_3 is produced?