# An excess of hydrogen reacts with 14.0 g of N_2. What volume of ammonia will be produced at STP?

May 3, 2016

$V = 33.6 L$

#### Explanation:

The reaction between hydrogen and nitrogen is the following:

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

Using the mass of nitrogen $m = 14.0 g$, we can first find the number of mole of ammonia that will be produced:

?molNH_3=14.0cancel(gN_2)xx(1cancel(molN_2))/(28.0cancel(gN_2))xx(3molNH_3)/(1cancel(molN_2))=1.50molNH_3

It is known that at STP, one mole of gas occupies a volume of $22.4 L$. This is called the molar volume.

Thus, the ammonia produced will occupy the following volume:

?L=1.50cancel(molNH_3)xx(22.4L)/(1cancel(molNH_3))=33.6 L