# Using the following reaction, if 1.75 moles of nitrogen gas are reacted with excess hydrogen gas, how many moles of NH_3 will be produced? N_2 + 3H_2 -> 2NH_3?

$\frac{n \left(N {H}_{3}\right)}{n \left({N}_{2}\right)} = \frac{2}{1}$
$n \left(N {H}_{3}\right) = 2 n \left({N}_{2}\right) = 2 \times 1.75 m o l = 3.5 m o l$