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)

Explanation

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 of electrone tranfer has occured.
Ampere-hour `=(9.888xx96500)/(3600)=2.65Ah`