Question

3.6 grammes of magnesium combined with 1.4 grammes of nitrogen. Calculate the simplest formula of the compound formed

Answer 1

Moles of $Mg=\frac{3.6}{24}=0.15mol$

Moles of $N_2=\frac{1.4}{28}=0.05mol$

As number of moles of $Mg$ reacted$=3\times$ number of moles of $N_2$ reacted

Hence the simplest formula of the compound formed is $Mg_3N_2$