# What is the volume of NH_3 produced in the following reaction when 3.0 L of N_2 reacts with 4.0 L of H_2?

Dec 2, 2015

${V}_{N {H}_{3}} = 2.7 L$

#### Explanation:

The balanced reaction is the following:

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \to 2 N {H}_{3} \left(g\right)$

We will assume that this reaction is happening at constant pressure and that the temperature is not changing through the entire process, therefore, we can say that:

$P V = n R T \implies V = {\underbrace{\frac{R T}{P}}}_{\textcolor{b l u e}{\text{constant}}} n \implies$$V \alpha \text{ n}$

Therefore, now we can use volume the same way that we use number of mole in stoichiometric calculations:

?LNH_3=4.0LH_2xx(2LNH_3)/(3LH_2)=2.7LNH_3

?LNH_3=3.0LN_2xx(2LNH_3)/(1LN_2)=6.0LNH_3

Since ${H}_{2}$ produces the smallest amount of $N {H}_{3}$ therefore, it is the limiting reactant and the volume of $N {H}_{3}$ produced is $2.7 L$.