Calcium phosphide $$(Ca_3P_2)$$ formed by reacting calcium orthophosphate $$(Ca_3(PO_4)_2)$$ with magnesium was hydrolysed by water. The evolved phosphine$$(PH_3)$$ was burnt in air to yield phosphorus pentoxide $$(P_2O_5)$$ How many grams of magnesium metaphosphate would obtained? If 19.2 g of magnesium were used for reducing calcium phosphate (At. wt Mg =24, P=31)
$$(Ca_3(PO_4)_2) +Mg \rightarrow Ca_3P_2 +MgO$$
$$Ca_3P_2 +H_2O \rightarrow Ca(OH)_2+ PH_3$$
$$PH_3 +O_2 \rightarrow P_2O_5 +H_2O$$
$$MgO + P_2O_5 \rightarrow Mg(PO_3)_2$$
Balanced chemical equation are:
$$Ca_3 (PO_4)_2 +8Mg \rightarrow Ca_3P_2+8MgO$$
$$Ca_3P_2+ 6H_2O \rightarrow 3Ca(OH)_2 +2PH_3$$
$$2PH_3 +4O_2 \rightarrow P_2O_5 + 3H_2O$$
$$MgO + P_2O_5 \rightarrow Mg (PO_3)_2$$
moles of magnesium used=0.8 moles
moles of $$MgO$$ formed =0.8 moles
moles of $$Ca_3P_2$$ formed 0.1 moles
moles of $$PH_3$$ formed=0.2 moles
moles of $$P_2O_5$$ formed =0.1 mole (limiting reagent)
Moles of $$Mg (PO_3)_2$$
=0.1 moles
mass of $$Mg (PO_3)_2$$
=18.2 grams