# N 2 ( g ) + 3 H 2 ( g ) ⟶ 2 N H 3 ( g ) In a certain reaction, 5.0 moles of hydrogen is reacted with excess nitrogen. To determine how many moles of ammonia are produced, what conversion factor should be used?

Apr 25, 2018

Here ${H}_{2}$ will act as the Limiting reagent because the ${N}_{2}$ is take in excess. See below :-

#### Explanation:

N_2 + 3H_2 ⟶ 2NH_3

On dividing by $3$ , this equation can be rewritten as :-

1/3N_2 + H_2 ⟶ 2/3NH_3

This equation explains that for $1$ mole of ${H}_{2}$ the number of moles of $N {H}_{3}$ formed will be $\frac{2}{3}$. (when ${N}_{2}$ is in excess)

So for $5$ moles of ${H}_{2}$ the number of moles of $N {H}_{3}$ formed $= \frac{2}{3} \times 5 = \frac{10}{3}$moles

Thus the number of moles of Ammonia formed $= \frac{10}{3}$