Question

Considering the limiting reactant, what is the volume of NO gas produced from 60.0L of ammonia gas and 50.0L of oxygen gas? (assume constant conditions)

Answer 1

The equation for this reaction is next:

$$4 \mathrm{NH_3} + 5 \mathrm{O_2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H_2O}$$

As you can see volume and mole ratio between $\mathrm{NH_3}$ and $\mathrm{O_2}$ is 4:5.

For example $40\,\mathrm{L}$ of $\mathrm{NH_3}$ reacts with $50\,\mathrm{L}$ of $\mathrm{O_2}$. In your case, where volume of ammonia is $60\,\mathrm{L}$ and volume of oxygen is $50\,\mathrm{L}$, the last one is definitely in shortage, so it is limiting reactant.

If you have $60\,\mathrm{L}$ of ammonia you need $\mathrm{X}\,\mathrm{L}$ of oxygen for its complete reaction:

$$60\,\mathrm{L} \quad \mathrm{x}$$ $$4 \mathrm{NH_3} + 5 \mathrm{O_2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H_2O}$$ $$4 \quad 5$$

$\mathrm{x} = 60 \times 5/4 = 75\,\mathrm{L}$ but you have only $50\,\mathrm{L}$, oxygen is limiting agent.