During discharging of lead storage acid battery following reaction takes place 
$P{b}_{\left(s\right)}+Pb{O}_{2\left(s\right)}+2H_{2}S{O}_4\to 2PbS{O}_{4\left(s\right)}+2H_{2}O$
If 2.5 amperes of current is drawn for 965 min H₂SO₄ consumed is

Q=ct
Q=2.5x965x60 colulombs
=144750 coulombs
No. of Faradays =$\frac{144750}{96500}=1.5$
No. of mole of $e^{-}=1.5$
2 moles of $H_{2}S{O}_4$ is consumed by exchange of  2mole of electrons.
$\therefore$ No. of moles of   consumed by exchange of 1.5mole of electrons = $\frac{1.5}{2}\times 2$=1.5