1.0 g of Mg is burnt in a closed vessel which contains 0.5 g of `O_2`.
(i) Which reactant is left in excess?
(ii) Find the weight of excess reactant.
(iii) How many millilitres of `0.5 N H_2 SO_4` will dissolve the residue in the vessel?

Explanation

`Mg+O_2 to MgO`
Equivalent of `Mg = (1)/(12) = 0.0833" "("eq. wt. of" Mg = ( "at. wt." )/( "valency") = ( 24)/(2) )`
Equivalent of oxygen `(0.5)/( 8) = 0.0625 " (eq. wt. of O = 8)"`
(i) Mg is in excess because its eq. is greater than that of oxygen.
(ii) Equivalent of Mg in excess `=0.0833 - 0.0625`
`= 0.0208`
`therefore` weight of Mg in excess `=` equivalent `xx` equivalent weight
`= 0.0208 xx 12 = 0.25 g`
(iii) The residue contains MgO and Mg which has not taken part in the reaction. Suppose `upsilon` mL of `H_2 SO_4` is required to dissolve the residue.
`therefore` m.e. of `H_2 SO_4` = m.e. of Mg + m.e. of MgO (not reacted)
`therefore 0.5 upsilon` = (eq. of Mg + eq. of MgO)
` = 1000` (eq. of Mg `+` eq. of O)
`=1000 (0.0208 + 0.0625)`
`= 83.3`
`therefore upsilon= 166.6` mL.