Question

Calcium phosphide `Ca_(3)P_(2)` formed by reacting magnesium with excess calcium orthophosphate `Ca_(s)(PO_(4))_(2)` was hydrolysed by excess water. The evolved. Phosphine `PH_(5)` was burnt in air to yield phosphrous pentoxide `(P_(2)O_(6))` . How many gram of magnesium metaphosphate would be obtain if 192 gram Mg were used (Atomic weight of Mg=24, P=31)
`Ca_(3)(PO_(4))_(2) + Mg to Ca_(3)P_(2) +_ MgO`
`Ca_(3)P_(2)+H_(2)O to Ca(OH)_(2) + PH_(3)`
`PH_(3) + O_(2) to P_(2)O_(5) + H_(2)O`
`MgO + P_(2)O_(5) to Mg(PO_(3))_(2)`
`" "` magnesium metaphosphate.

Answer 1

POAC on Mg
`1xx n_("Mg")=1xxn_("Mg"(PO_(3))_(2))`
`(192)/(24)=n_("Mg"(PO_(3))_(2))`
`W_(Mg(PO_(3))_(2))=(192)/(24)xx182gm =1456 gm`