Question

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

Answer 1

Firstly, fully balance the chemical equation.
`N_2 + 3H_2 => 2NH_3`

Use this formula `n = 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 = 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 `xx` 2 = 28.0 g/mol

So,
`n` = `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 `NH_3`. This is done using the mole ratio.
The mole ratio between `N_2`:`NH_3` is `1:2`.

This means if 1 mole of `N_2` gives you 2 moles of `NH_3`,
then 0.6607142857 moles of `N_2` will give you:
`0.6607142857 -: 1 xx 2` = 1.32 moles of `NH_3`.

Therefore, 1.32 moles of `NH_3` are produced when 18.5 grams of `N_2` gas reacts with excess `H_2` gas.