During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139 g/mL. Supphuric acid of density 1.294 g/mL is 39% H_(2)SO_(4 )by weight and that of density 1.139 g/mL is 20% H_(2)SO_(4)by weight. The battery hold 3.5 L of the acid and the volume remained particlcally constant during the discharge.Calculate the number of ampere-hours for which the battery must have been used. The charging and discharging reactions are Pb+SO_(4)^(2-) = PbSO_(4) +2e^(-) (charging) PbO_(2)+ 4H^(+) + So+(4)^(2-) = PbSo_(4) + 2H_(2)O (discharging)

During the discharge of a lead storage battery, the density of sulphur

1.	During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139 g/mL. Supphuric acid of density 1.294 g/mL is 39% H_(2)SO_(4 )by weight and that of density 1.139 g/mL is 20% H_(2)SO_(4)by weight. The battery hold 3.5 L of the acid and the volume remained particlcally constant during the discharge.Calculate the number of ampere-hours for which the battery must have been used. The charging and discharging reactions are Pb+SO_(4)^(2-) = PbSO_(4) +2e^(-) (charging) PbO_(2)+ 4H^(+) + So+(4)^(2-) = PbSo_(4) + 2H_(2)O (discharging)
Answer» Solution :For 1.0 L `H_2SO_4`: INITIAL mass of `H_2SO_4=1294xx39/100=504.66g,` FINAL mass of `H_2SO_4=1139xx20/100=227.80g` `implies H_2SO_4` Consumed/litres `=504.66-227.80=276 g` `implies` Total `H_2SO_4` USED up `=276.86xx3.5=969.01 g=(969.01)/(98) m ol= 9.888 m ol` Q 1 mole of `H_2SO_4` is associated with transfer of .0 mole of electrons, total of 9.888 mole ofelectrone tranfer has occured. Ampere-hour `=(9.888xx96500)/(3600)=2.65Ah`