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?

1.0 g of Mg is burnt in a closed vessel which contains 0.5 g of O_2.

1.	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?
Answer» Solution :`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.