Question

How many molecules (not moles) of NH3 are produced from 3.20×10−4 g of H2?

Express your answer numerically as the number of molecules.

Answer 1

First solution

$$3H_2 + N_2 = 2NH_3$$

The stoichiometric ratio of Hydrogen H₂ to Ammonia NH_3:

$$3 : 2$$

$$n(H_2) = {m \over M} = {3.2×10^{-4} \over 2} = 1.6×10^{-4} moles$$

Based on the ratio, the amount of substance Ammonia NH_3:

$$n(NH_3) = {1.6×10^{-4} \over 1.5} = 1.067×10^{-4} moles$$

$$N(NH_3) = N_a×n = 6.02×10^{23}×1.067×10^{-4} = 6.42×10^{19} molecules$$

Second solution

$$n(H) = {m \over M} = {3.2×10^{-4} \over 1} = 3.2×10^{-4} moles$$

If one mole of ammonia contains three moles of Hydrogen atoms, then:

$$n(NH3) = {n(H) \over 3} = {3.2×10^{-4} \over 3} = 1.067×10^{-4} moles$$

$$N(NH_3) = N_a×n = 6.02×10^{23}×1.067×10^{-4} = 6.42×10^{19} molecules$$