0.5 moles of $H_{2}S{O}_4$ is mixed with 0.2 moles of $Ca(OH{)}_2.$ The maximum number of moles of $CaS{O}_4$ formed is:
Reaction: $H_{2}S{O}_4+Ca(OH{)}_2\to CaS{O}_4+2H_{2}O$

The correct option is A 0.2 mol
The given balanced equation is, $H_{2}S{O}_4+Ca(OH{)}_2\to CaS{O}_4+2H_{2}O$
1 mol of $H_{2}S{O}_4$ reacts with 1 mol of $Ca(OH{)}_2$ to give 1 mol of $CaS{O}_4.$
The amount of $H_{2}S{O}_4=0.5$ mol
The amount of $Ca(OH{)}_2=0.2$ mol
Thus $Ca(OH{)}_2$ is the limiting reagent.
Hence, the maximum number of moles of $CaS{O}_4$ formed in the reaction is 0.2 mol.