# Question #e0a35

Feb 11, 2018

There's a little shortcut here where you don't need to use:

$P V = n R T$

Since pressure and temperature remain constant, and you know there are half as many moles of ${N}_{2}$ as there are ${H}_{2} O$, then you can determine that the volume of ${N}_{2}$ produced is half the volume of ${H}_{2} O$ This only works with constant temperature and pressure with knowledge of the balanced chemical equation.

So you have 25 liters of ${N}_{2}$

If you actually want to calculate it:

One mole of gas at STP takes up about 22.4 L of space.

$\frac{50 L}{22.4 \left(\frac{L}{m o l}\right)} = 2.23 m o l \left({H}_{2} O\right)$

$2.23 m o l \left({H}_{2} O\right) \cdot \frac{1 m o l \left({N}_{2}\right)}{2 m o l \left({H}_{2} O\right)} = 1.116 m o l \left({N}_{2}\right)$

STP means $T = 273.15 K$ and $P = 1.0 a t m$

$R$ is the gas constant and for atm it is 0.08206

$V = \frac{n R T}{P}$

$V = \frac{\left(1.12 m o l {N}_{2}\right) \left(0.08206\right) \left(273.15 K\right)}{1.0 a t m} = 25.0 L \left({N}_{2}\right)$