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.
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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`
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