How many moles of $P_4{O}_6$ and $P_4{O}_{10}$ will be produced by the combustion of $12.4 \mathrm{gm}$ of phosphorus in  $12.8 \mathrm{gm}$ of $O_2$, leaving no $P_4$ or $O_2$?
[Atomic wt. $\mathrm{P}=31$]

No. of moles of ${\mathrm{P}}_4=\frac{12.4}{124}=0.1$

No. of moles of ${\mathrm{O}}_2=\frac{12.8}{32}=0.4$

Let moles of $P_4{O}_6$ and $P_4{O}_{10}$ be $n_1\text{ and }{n}_2$ respectively.

POAC on 'P'$4\times 0.1=n_1\times 4+n_2\times 4$

POAC on $O2\times 0.4=n_1\times 6+n_2\times 10$

$n_1=n_2=0.05 \mathrm{mol}$