Question

A sample of magnisium metal containing some `MgO` as impurity was dissolved in 125 " mL of " 0.1 N `H_2SO_4`. The volume of `H_2` evolved at `27.3^@C` and 1 atm was 120.1 mL. The resulting solution was found to be 0.02 N with respect to `H_2SO_4`. Calculate (i) the weight of sample dissolved and (ii) the percentage by weight of Mg in the sample. Neglect any change in the volume of the solution (atomic weight of `Mg=24.3`).

Answer 1

Volume of `H_2` at `STP=(120.1xx273)/(300.3)=109.2mL`
`Mg+H_2SO_4toMgSO_4+H_2`
`MgO+H_2SO_4toMgSO_4+H_2O`
Weight of Mg`=(24.3)/(22400)xx109.2=0.1185g`
`" Eq of "Mg=(0.1185)/(12.15)=0.009753`
`" Eq of "H_2SO_4(t otal)=(125xx0.1)/(1000)=0.0125`
Excess of `H_2SO_4=(125xx0.02)/(1000)=0.0025`
Excess of `H_2SO_4=(125xx0.2)/(1000)=0.0025`
`H_2SO_4` used`=0.0125-0.0025=0.01`
`=g" Eq of "Mg+g" Eq of "MgO`
" Eq of "MgO`=` Total `Eq-" Eq of "Mg`
`=0.01-0.009753`
`0.000247 " Eq of "MgO`
`=0.00247xx(40.3)/(2)g of MgO`
`=0.005g`
Total weight `=` Weight of `Mg+` weight of `MgO`
`=0.1185+0.005=0.1235g`
`% of Mg=(0.1185)/(0.1235)xx100=95.5%`