# For the reaction represented by the equation N_2 + 3H_2 -> 2NH_3, how many moles of nitrogen are required to produce 18 mol of ammonia?

Jan 26, 2016

$9 m o l {N}_{2}$

#### Explanation:

The reaction is:

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \to 2 H {N}_{3} \left(g\right)$

The molar ratio between Nitrogen and ammonia is $1 : 3$, therefore, to produce 18 moles of ammonia, we will need:

?molN_2=18cancel(molNH_3)xx(1molN_2)/(2cancel(molNH_3))=9molN_2

May 1, 2016

Compare the mole ratio between $N {H}_{3}$ and ${N}_{2}$.
The mole ratio between $N {H}_{3} : {N}_{2}$ is $2 : 1$.

If 2 moles of $N {H}_{3}$ gives you 1 mole of ${N}_{2}$,

then 18 moles of $N {H}_{3}$ should give you:
[18 moles$\div$ 2 $\times$ 1] = 9 moles of ${N}_{2}$

Therefore 9 moles of nitrogen are required to produce 18 moles of ammonia.