Molar mass of CaCO3 is 100g∕mol. 10g of 100% pure limestone =10g of CaCO3 So, 10g of 90% pure limestone =9g of CaCO3 This implies that, 9g of CaCO3=0.09‌moleof CaCO3 The chemical reaction occurs during the heating of CaCO3 is written as,
CaCO3
1‌mol
Δ
––––––🢖
CaO+
CO2(g)
1‌mol
By stoichiometry, 1 mole of CaCO3 yields 1 mole of CO2(g) or 22.4L of CO2 The volume of CO2 contained in 0.09‌mole of CaCO3 iscalculated as, 0.09 mole of CaCO3=0.09×22.4L‌of‌CO2 =2.016L of CO2