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?

1 Answer
Dr. Hayek
Dec 2, 2015

V_(NH_3)=2.7L

Explanation:

The balanced reaction is the following:

N_2(g)+3H_2(g)->2NH_3(g)

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:

PV=nRT=>V=underbrace((RT)/P)_(color(blue)("constant"))n=>Valpha" 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 NH_3 therefore, it is the limiting reactant and the volume of NH_3 produced is 2.7L.

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