1.2gm of Mg is burnt in a closed vessel which contains 0.5gm of O₂.Millilitres of 0.25M H₂So₄ reqired to dissolve the residue in the vessel is

1.12.5ml

2.175ml

3.200ml

4.250ml

2 Mg + O2 → 2 MgO

(1 g Mg) / (24.30506 g Mg/mol) = 0.0411 mol Mg
(0.5 g O2) / (31.99886 g O2/mol) = 0.0156 mol O2

0.0156 mole of O2 would react completely with 0.0156 x (2/1) = 0.0312 mole of Mg, but there is more Mg present than that, so Mg is in excess.

((0.0411 mol Mg initially) - (0.0312 mol Mg reacted)) x (24.30506 g Mg/mol) = 0.24 g Mg left over