# What volume of "NH"_3 at STP is produced if 25.0 "g" of "N"_2 is reacted with an excess of "H"_2 in the reaction "N"_2 + 3"H"_2 -> 2"NH"_3?

Mar 16, 2016

$40.0 \text{L}$

#### Explanation:

$1 \text{mol}$ of ${\text{N}}_{2}$ has mass of $28.0 \text{g}$

Therefore, $25.0 \text{g}$ of ${\text{N}}_{2}$ in moles is

frac{25.0"g"}{28.0 "g/mol"} = 0.893 "mol"

Each molecule of ${\text{N}}_{2}$ will produce 2 molecules of ${\text{NH}}_{3}$.

Therefore, the 2 times the amount of ${\text{NH}}_{3}$ will be produced.

$2 \times 0.893 \text{mol" = 1.79 "mol}$

One mole of an ideal gas has a volume of $22.4 \text{L}$ at STP.

Therefore, the volume occupied by $1.79 \text{mol}$ of ${\text{NH}}_{3}$ is

$1.79 \text{mol" xx 22.4 "L/mol" = 40.0 "L}$

That's it. But bear in mind that you need to somehow remove the excess ${\text{H}}_{2}$ before the container will have a volume of $40.0 \text{L}$.