An impure solid sample of CaCO₃ weighing 50.0 g and having 80 % purity by mass is allowed to react with 450 mL of 2.0 M aqueous HCl solution. After the reaction is complete, what is the mass of the excess reactant left unreacted? [Given: Molar atomic mass in g mol ⁻¹ : Ca = 40 , C = 12 , O = 16 , H = 1 , Cl = 35.5 ]

Step-by-step solution

The balanced chemical equation is: CaCO₃(s) + 2HCl(aq) CaCl₂(aq) + CO₂(g) + H₂O(l) Mass of pure CaCO₃ in the sample = 50.0 80 100 = 40.0 g . Molar mass of CaCO₃ = 100 g mol ⁻¹ . Moles of CaCO₃ = 40.0 100 = 0.4 mol . Moles of HCl available = Molarity Volume (in L) = 2.0 0.450 = 0.9 mol . From the stoichiometry, 1 mol of CaCO₃ requires 2 mol of HCl . Moles of HCl required to react completely with 0.4 mol of CaCO₃ = 0.4 2 = 0.8 mol . Since the available HCl ( 0.9 mol ) is greater than the required HCl ( 0.8 mol ), CaC