0.56 g of a lime stone was treated with oxalic acid to give C a C 2 O 4 . The precipitate decolourized 45 mL of 0 .2 N KMnO 4 in acid medium. Calculate % of C a O in the stone.

Step-by-step solution

The given reaction is CaCO 3 lime stone → H 2 C 2 O 4 CaC 2 O 4 → KMnO 4 decolourizes The redox changes are : C 3 + 2 → 2 C 4 + + 2 e M n 7 + + 5 e → M n 2 + Meq . of CaO = Meq . of CaCO 3 = Meq . of H 2 C 2 O 4 = Meq . of CaC 2 O 4 = Meq . of KMnO 4 Hence Meq .   of   CaO   =   Meq .   of   KMnO 4 w / ( 56 / 2 ) × 1000 = 45 × 0.2 So wt . of CaO = 0 .252 g Here % of C a O in lime stone = 0.252 0.256 × 100 = 45 %