# How many moles of NH_3 are produced when 18.5 grams of N_2 gas reacts with excess H_2 gas?

Dec 11, 2016

Firstly, fully balance the chemical equation.
${N}_{2} + 3 {H}_{2} \implies 2 N {H}_{3}$

Use this formula $n = \frac{m}{M}$ where:
$n$ = number of moles
$m$ = mass of substance
$M$ = molar mass

Secondly, you need to find out how many moles of ${N}_{2}$ there are.
Use that $n = \frac{m}{M}$ formula to do that.
$n$ = ?
$m$ = 18.5 grams
$M$ = ?

In order to calculate what is $n$, you need to find the $M$ of ${N}_{2}$ first.
The molar mass can be determined right away by looking straight into the periodic table.
Molar mass of ${N}_{2}$ is = 14.0 g/mol $\times$ 2 = 28.0 g/mol

So,
$n$ = $\frac{18.5}{28.0}$ = 0.6607142857 moles [Don't round it off yet or you'll get inaccurate final answer in the end!]
$m$ = 18.5 grams
$M$ = 28.0 g/mol

Third step, find out what the $n$ is for $N {H}_{3}$. This is done using the mole ratio.
The mole ratio between ${N}_{2}$:$N {H}_{3}$ is $1 : 2$.

This means if 1 mole of ${N}_{2}$ gives you 2 moles of $N {H}_{3}$,
then 0.6607142857 moles of ${N}_{2}$ will give you:
$0.6607142857 \div 1 \times 2$ = 1.32 moles of $N {H}_{3}$.

Therefore, 1.32 moles of $N {H}_{3}$ are produced when 18.5 grams of ${N}_{2}$ gas reacts with excess ${H}_{2}$ gas.