# Given the equation:N_2+3H_2 rarr 2NH_3, what volume of NH_3 at STP is produced if 25 grams of N_2 is reacted with an excess of H_2?

Jan 27, 2016

$V = 40 L$

#### Explanation:

One thing I think it's interesting to note is that you'll never be able to have that reaction happen in the standard temperature and pressure; you can, however, have the resulting ammonia in the STP.

Anyhow, ${N}_{2}$ has molar mass $14 \cdot 2 = 28$, so for 25 g we have

${n}_{{N}_{2}} = \frac{m}{M M} = \frac{25}{28} m o l$

Every one mol of nitrogen makes two moles of ammonia, so we have

${n}_{N {H}_{3}} = 2 {n}_{{N}_{2}} = 2 \cdot \frac{25}{28} = \frac{25}{14} m o l$

Since we know that at STP 1 mol of gas has 22.4 L (this can easily be checked by plugging the appropriate values in $p V = n R T$) We have

${V}_{N {H}_{3}} = 22.4 \cdot \frac{25}{14} = 40 L$