What weights of $P_4{O}_6\mathit{ }and\mathit{ }{P}_4{O}_{10}$ will be produced by the combustion of  $of\mathit{ }31g\mathit{ }of\mathit{ }{P}_4\mathit{ }in\mathit{ }32g$ of oxygen leaving $no\mathit{ }{P}_{4 }and\mathit{ }{O}_2$ ?

$\text{ Percentage of }{\mathrm{P}}_4{\mathrm{O}}_6=\frac{(31\times 4)\times 100}{(31\times 4+32\times 6)}=56.3437\mathrm{\%}\mathrm{P}$

$\text{ Percentage of }{\mathrm{P}}_4{\mathrm{O}}_{10}={43.6420}^{\circ }\mathrm{P}$

$31\mathrm{gP}=31/(31+32)=49.2063\mathrm{\%}\mathrm{P}$

So,

$\begin{array}{r}100\mathrm{\%}{\mathrm{P}}_4{\mathrm{O}}_6=56.3437\mathrm{\%}\\ 2\mathrm{\%}{\mathrm{P}}_4{\mathrm{O}}_6=49.2063\mathrm{\%}\\ 0\mathrm{\%}{\mathrm{P}}_4{\mathrm{O}}_6=43.6420\mathrm{\%}\end{array}$

Find the value of z

$\begin{array}{r}\frac{z}{100}=(49.2063-43.6420)(56.3437-43.6420)\\ z=43.808\mathrm{\%}{\mathrm{P}}_4{\mathrm{O}}_6\end{array}$

By the law of the Conservation of Matter, the total mass of ${\mathrm{P}}_4{\mathrm{O}}_6\text{ and }{\mathrm{P}}_4{\mathrm{O}}_{10}=31\mathrm{g}+32\mathrm{g}=63\mathrm{g}$

$43.808\mathrm{\%}\text{ of }63\mathrm{g}=27.6{\mathrm{gP}}_4{\mathrm{O}}_6$

$63\mathrm{g}-27.6\mathrm{g}=35.4{\mathrm{gP}}_4{\mathrm{O}}_{10}$