CO2 This implies that 2 moles of KOH are required to neutralize 1mole of CO2. The molar mass of KOH given is 56 g. The mass of KOH will be 112(56×2) since 2 moles are reacting. 112g of NaOH will neutralize 100g of CaCO3 So 28g of KOH will neutalize =
100×28
112
=25g of CaCO3 Now, 60 g of impure CaCO3, has 25g of pure CaCO3. So, the percentage purity present in 100 g of impure CaCO3 iscalculated as follows: Percentage purity =