Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below:
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.

Calculation:

The number of moles of HCl taken = MV/1000

The number of moles of HCl taken = 0.76*250/1000

The number of moles of HCl taken = 0.19

Number of moles of CaCO3 = Mass/Molar mass

Number of moles of CaCO3 = 1000/100

Number of moles of CaCO3 = 10

1. If CaCO3 is completely consumed,

1 mol of CaCO3 = 1 mol CaCl2

10 mol CaCO3 = 10mol CaCl2

1. If HCl is completely consumed,

2 mol HCl = 1 mol CaCl2

0.19mol HCl = ½ × 0.19 mol CaCl2

0.19mol HCl = 0.095 mol CaCl2

HCl will be the limiting reagent

Hence, the number of moles of CaCl2 formed will be 0.095mol