# Question #71e84

Jul 27, 2017

${V}_{N {H}_{3}} \approx 20. L$

#### Explanation:

This is a stoichiometry-gas law problem.

${N}_{2} + 3 {H}_{2} \to 2 N {H}_{3}$

$\frac{10 L}{22.4 L} = 0.446 m o l \left[{N}_{2}\right]$
$5 g \cdot \frac{2.02 g}{H} _ 2 = 2.48 m o l \left[{H}_{2}\right]$

Based on stoichiometry, you can see that ${N}_{2}$ will be the limiting reactant, so let's calculate the amount at STP.

$1.0 a t m \cdot 10.0 L = n \cdot \frac{0.08206 L \cdot a t m}{m o l \cdot K} \cdot 273.15 K = n = 0.446$

Well, that was a waste of time, but it's nice to double check that they are the same, anyways:

$0.446 m o l \cdot \frac{2 N {H}_{3}}{{N}_{2}} \approx 0.892 m o l \left[N {H}_{3}\right] = n$

$1.0 \cdot V = 0.892 \cdot \frac{0.08206 L \cdot a t m}{m o l \cdot K} \cdot 273.15 K$
$\therefore V \approx 20. L$