Excess of calcium orthophosphate is reacted with magnesium to from calcium phosphide `(Ca_(3)P_(2))` along with magnesium oxide. Calcium phosphide on reacting with excess of water liberate phosphine gas `(PH_(3))` along with calcium hydroxide. Phosphine is burnt in excess of oxygen to from `P_(2)O_(5)` along with water. oxides of magnesium & phosphorous react to give magnesium metaphosphate. calculate grams of magnesium metaphosphate obtained if 1.92 gm of magnesium is taken .

Answer

2

Explanation

`Ca_(3)(PO_(4))_(2)+underset(0.08)(8 Mg) to underset("0.01 Mole")(Ca_(3)P_(2))+underset(0.08)(8 MgO)`
`underset(0.01)(Ca_(3)P_(2))+6H_(2)Oto underset(0.02) (2PH_(3))+3Ca(OH)_(2)`
`underset(0.02)(2PH_(3))+4O_(2) to underset(0.01)(P_(2)O_(5))+3H_(2)O)`
`underset(MgO)(0.08) +underset(0.01)(P_(2)O_(5))to underset("0.01 Mole" ~~ 1.82 gm ~~ 2gm)(MgP_(2)O_(6))`