Answer
0.235 mole of $NH_3$
Work Step by Step
1. Determine the molar mass of this compound ($NH_3$), and setup the conversion factors:
Molar mass :
$N: 14.01g $
$H: 1.008g * 3= 3.024g $
14.01g + 3.024g = 17.03g
$ \frac{1 \space mole \space (NH_3)}{ 17.03 \space g \space (NH_3)}$ and $ \frac{ 17.03 \space g \space (NH_3)}{1 \space mole \space (NH_3)}$
2. Calculate the number of moles $(NH_3)$
$ 4.00g \times \frac{1 mole}{ 17.03g} = 0.235 mole$
