# Question #7ea2e

The interacting ratio of number of moles of $N {H}_{3} \mathmr{and} {O}_{2}$ as per above equation is $4 : 5 = \frac{4}{5} : 1 = 0.8 : 1$
So if the 1 mole of each reactant here is made to react, ${O}_{2}$ will be fully consumed reacting with $0.8$ mole $N {H}_{3}$ and as a result $0.2$ mole of $N {H}_{3}$ only will remain as residue.