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The mass of calcium carbonate required to react with 1L of 1 m HCl to liberate 11.2L of CO2 IS |
| CaCO3 + 2HCl -> CaCl2 + CO2 +H2O 1 L of 1 molar HCl solution contain 1 mol HCl. So, According to the above equation, 1/2 mol CaCO3 is needed to liberate 11.2 l of Co2. Molar mass of CaCO3 = 100 g So, mass of 1/2 mol CaCO3 = 50 g |
