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.

This is a Long Answer Type Questions as classified in NCERT Exemplar The volume of HCl solution is 250 mL and its molarity is 0.76M. The number of moles of HCl as follows, Moles of HCl Molarity Volume (in L) = 0.76M 0.250 L = 0.19 mol The molar mass of CaCO3 is 100 g / gQl and the mass of CaCO3 is given as 1000 g The number of moles of CaCO3 is calculated as Moles of CaCO3 = M a s / M o l a r m a s = 1000 g / 100 g / m o l = 10 mol According to the given reaction, 1 mole of CaCO3 requires 2 moles of HCl. So, the required number of moles of HCl for 10 moles of CaCO3 is calculated as Moles of HCl = 2mol of HCl /1mol of CaCO2 * 10 mol of CaCO3 = 20 mol So, the required number of moles of HCl is 20 mol but only 0.19 mol are given. So, HCl is a limiting reagent. So, the amount of calcium chloride formed is depend upon the limiting reagent, that is, the amount of HCl available. According to the reaction, 2 moles of HCl gives 1 mol of CaCl2 Sg 2 ., the number of moles of calcium chloride produced by 0.19 mol of HCl as follows, Moles of CaCl2 = 1 mol of CaCl2 / 2 mol of HCl x 0.19 mol of HCl = 0.095 mol The molar mass of calcium chloride is 111 g / mol. So, its mass is calculated as Mass of CaCl2 = Moles x Molar mass = 0.095 mol x 111 g / mol = 10.54 g