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
Dec 2, 2015

Answer:

#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#.