100g of CaCO3 is treated with 1 liter of 1 NHCl. What would be the weight of CO2 liberated after the completion of the reaction

Explanation

Here's how to solve the problem:Set up the balanced chemical equation:CaCO3 (s) + 2 HCl (aq) -> CaCl2 (aq) + CO2 (g) + H2O (l)Calculate the number of moles of HCl:Volume of HCl = 1 LConcentration of HCl = 1 N (1 N = 1 mol/L)Number of moles of HCl = Volume x Concentration = 1 L x 1 mol/L = 1 molDetermine the limiting reactant:From the balanced equation, 1 mol of CaCO3 reacts with 2 mol of HCl.We have 1 mol of HCl and 100 g of CaCO3.Molar mass of CaCO3 ≈ 100 g/mol.Number of moles of CaCO3 = Mass / Molar mass = 100 g / 100 g/mol = 1 mol.Since we have the same number of moles of CaCO3 and HCl (1 mol each), neither is in excess. Therefore, both will react completely, and HCl is the limiting reactant.Calculate the number of moles of CO2 produced:From the balanced equation, 1 mol of HCl produces 1 mol of CO2.Since 1 mol of HCl is the limiting reactant, it will produce 1 mol of CO2.Calculate the mass of CO2 produced:Molar mass of CO2 ≈ 44 g/molMass of CO2 = Number of moles x Molar mass = 1 mol x 44 g/mol = 44 gTherefore, the weight of CO2 liberated after the completion of the reaction is 22 g (half of the 44 g produced because only 1 mol of HCl reacted, utilizing half the CaCO3 and producing half the CO2).