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.
No: of moles of HCl taken = MV/1000 = 0.76*250/1000 = 0.19
No: of moles of CaCO3 = Mass/Molar mass = 1000/100 = 10
1. When CaCO3 is completely consumed
1 mol of CaCO3 = 1 mol CaCl2
10 mol CaCO3 = 10mol CaCl2
2. When HCl is completely consumed.
2 mol HCl = 1 mol CaCl2
0.19mol HCl = ½ × 0.19mol CaCl2 = 0.095 mol CaCl2
HCl will be the limiting reagent and the number of moles of CaCl2 formed will be 0.095mol