Excess of calcium orthophosphate ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$ is reacted with magnesium to form calcium phosphide $\left({\mathrm{Ca}}_3{\mathrm{P}}_2\right)$ with magnesium oxide. Calcium phosphide on reacting with an excess of water liberate phosphine gas (${\mathrm{PH}}_3$) along with calcium hydroxide. The evolved phosphine $\left({\mathrm{PH}}_3\right)$ was burnt in the air (excess of oxygen ) to yield phosphorus pentoxide $\left({\mathrm{P}}_2{\mathrm{O}}_5\right)$ along with water. How many grams of magnesium metaphosphate would be obtained, if $19.2\mathrm{g}$ of magnesium were used for reducing calcium phosphate: 

( Atomic wt. of $\mathrm{Mg}=24, \mathrm{P}=31$ )

${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+ 8\mathrm{Mg}⟶{\mathrm{Ca}}_3{\mathrm{P}}_2+ 8\mathrm{MgO}$
${\mathrm{Ca}}_3{\mathrm{P}}_2+{\mathrm{H}}_2\mathrm{O}⟶\mathrm{Ca}(\mathrm{OH}{)}_2+{\mathrm{PH}}_3$
${\mathrm{PH}}_3+{\mathrm{O}}_2⟶{\mathrm{P}}_2{\mathrm{O}}_5+{\mathrm{H}}_2\mathrm{O}$
$\mathrm{MgO}+{\mathrm{P}}_2{\mathrm{O}}_5⟶\mathrm{Mg}{\left({\mathrm{PO}}_3\right)}_2$

Magnesium metaphosphate