Calcuim phosphide `(Ca_3P_2)` formed by reacting calcuim orthophosphate `(Ca_3(PO_4)_2)` with magnesium was hydrolsed 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 toCa_3P_2+MgO`
`Ca_3P_2+H_2OtoCa(OH)_2+PH_3`
`PH_3+O_2toP_2O_5+H_2O`
`MgO+P_2O_5tounderset("magnesium metaphosphate")(Mg(PO_3)_2)`

Answer

18 gram

Explanation

Balance chemical equations are :
`Ca_3(PO_4)_2+8Mgt OCa_3P_2+8MgO`
`Ca_3P_2+6H_2Oto3Ca(OH)_2+2PH_2`
`2PH_3+4O_2toP_2O_5+3H_2O`
`MgO+P_2O_5toMg(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 moles (limiting reagent)
moles of `Mg(PO_3)_2=0.1` moles
mass of `Mg(PO_3)_2=18.2` gram