Question

5. 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(1)
What mass of CaCl2 will be formed when 250 ml of 0.76M HCI reacts with 1000 gram of CaCO3?
Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.​

Answer 1

Number of moles of HCl=250 mL× 0.76 M/1000 = 0.19 mole

Mass of CaCO3 = 1000 g

Number of moles of CaCO3 = 1000 g/100 g = 10 mole

According to given equation 1 mole of CaCO3(s) requires 2 mole of HCl(aq). Hence, for the reaction of 10 mole of CaCO3(s) number of moles of HCl required would be :

10 mole CaCO3 × 2 mole HCl(aq)/1 mole CaCO3(s) = 20 mole HCl(aq)

But we have only 0.19 mole HCl(aq), hence HCl(aq) is limiting reagent. So, amount of CaCl2 formed will depend on the amount of HCl available. Since, 2 moles of HCl(aq) forms 1 mole of CaCl2, therefore, 0.19 mole of HCl(aq) would give:

0.19 mole HCl(aq) × 1 mole CaCl2(aq) /2 moles HCl(aq) = 0.095 mol of CaCl2

or 0.095 × molar mass of CaCl2 =0.095×111 = 10.54g