Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below: CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.
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Number of moles of HCl = 250 mL × = 0.19 mol 1000 Mass of CaCO3 = 1000 g 1000 g Number of moles of CaCO3 = = 10 mol 100 g According to given equation 1 mol of CaCO3 (s) requires 2 mol of HCl (aq). Hence, for the reaction of 10 mol of CaCO3 (s) number of moles of HCl required would be: 2 mol HCl (aq) 10 mol CaCO3 × 1 mol CaCO3 (s) = 20 mol HCl (aq) But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. So amount of CaCl2 formed will depend on the amount of HCl available. Since, 2 mol HCl (aq) forms 1 mol of CaCl2, therefore, 0.19 mol of HCl (aq) would give: 1 mol CaCl 2 (aq) 0.19 mol HCl (aq) × = 0.095 mol 2 mol HCl (aq) or 0.095 × molar mass of CaCl2 = 0.095 × 111 = 10.54 g