A 0.200M KOH solutions is electrolysed for 1.5 h using a current of 8.00 A. How many moles of O₂ were produced at the anode?

$\mathrm{Q}=\mathrm{It}$
Where q = charge, I = current, t = time
Charge $=8\mathrm{A}\times 1.5\times 60\times 60=43200\mathrm{C}$
$4\times 96500C$ produces $O_2$= 1mol
Therefore, 43200C will produce O₂=43200 /4×96500=0.112mol