Here, we have 1/2 moles of koh. 2 mol of KOH neutrailse 1 mol of Co2. So there is 1/4 mol of Co2.
Mol mass of CaCo3 is 100g, and 1 mol CaCO3 produces 1 mol of Co2. So to produce 1/4 mol of Co2, we need 1/4 mol = 25 gm of Caco3. So percentage of purity is 25/60 x 100 = 41.66%