1g of Mg is burnt in a closed vessel which contains `0.5g of O_(2)`
(i) Which reactants is left in excess
(ii) Find the mass of the excess reactant.

Explanation

The balanced equation is
`underset("=300g")underset(3(40+12+48))underset("3mol")(3CaCO_(3))+underset(=196g)underset(2(3+3+64))underset("2mol")(2H_(3)PO_(3)) to underset(=310g)underset((3xx40+2xx64))underset("1 mol")(Ca_(3)(PO_(4))_(2))+3CO_(2)+3H_(2)underset(=48g)underset(2xx24)(2Mg)+underset(=32g)underset(2xx16)(O_(2)) to underset(=80g)underset(2(24+16))(2MgO) `
48g of Mg require oxygen =32g
1 g of Mg requires oxygen =`(32)/(48)=0.66g`
But only 0.5g oxygen is available. Hence `O_(2)` is a limiting agent and a part fo magnesium will not burn.
Magnesium will be left in excess.
(ii) `32" g of " O_(2)` react with magnesium =48g
0.5g of `O_(2)` will react with magnesium =`(48)/(32)xx0.5=0.75g`
Hence, the mass of excess magnesium=`(1.0-0.75)=0.25g`