Calcium phoshide formed by reacting calcium orthophosphate with `Mg` was hydrolysed by water. The phosphine evolved was burnt in air to yield `P_2O_5`. How many litres of air, containg `21%` by volume of `O_2`, was required to burn `PH_3` and how many grams of magnesium metaphosphate would be obtained if 204 g of Mg was was used for reducing calcium phosphide and the volumes of the gases were measured at STP?

Explanation

`Ca(PO_4)_2+8"Mg" toCa_2P_2+8MgO` ..(i)
`Ca_3P_2+6H_2Oto3Ca(OH)_2+2PH_3` ..(ii)
`2PH_3+4O_2toP_2O_5+3H_2O` …(iii)
`P_2O_5+MgOtoMg(PO_3)_2` ..(iv)
From Equation (i):
`8 mol` of `Mg-=1 mol` of `Ca_3P_2-=8 " mol of "MgO`
`(2.4)/(24)=0.1 mol` of `Mg-=(0.1)/(8) mol` of `Ca_3P_2`
`-=0.1 mol` of `MgO`
From equation (ii):
`1 mol` of `Mg-=1 mol` of `Ca_3P_2-=mol` of `PH_3`
From equation (iii):
`2 mol` of `PH_3-=4mol` of `O_2-=1 mol` of `P_2O_5`
`(0.1)/(8) mol` of `PH_3-=(4)/(2)xx(0.1)/(4) mol` of `O_2`
`-=(1)/(2)((0.1)/(4))mol` of `O_2`
`-=00.5xx22.4` of `O_2` at `STO=1.121`
`-=(100)/(21)xx1.12L` of air
From equation (iv):
1 " mol of "`MgO-= 1 mol` of `P_2O_5` (it shows MgO is in excess)
`0.1` " mol of "`MgO -=(0.1)/(8)mol`
Weight of metaphosphate formed `-=(0.1)/(8)xx182=2.275`g