Question

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`?

Answer 1

`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`.