Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below: CaCO 3 (s) + 2HCl (aq)  → CaCl 2 (aq) + CO 2 (g) + H 2 O(l) What mass of CaCl 2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO 3 ? Name the limiting reagent. Calculate the number of moles of CaCl 2 formed in the reaction.

Question

Philoid · Accepted Answer

We need to compute (i) What mass of CaCl₂ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃?

(ii) Name the limiting reagent

(iii) Number of moles of CaCl₂ formed in the reaction.

Number of moles of HCl are not known. 0.76M in a 250mL solution gives 0.76M x 0.25L = 0.19 mol of HCl.

Molecular mass of CaCO₃ is 100. Number of moles present in 1000g of CaCO₃ is 1000/100 = 10 mol.

The limiting reagent is defined as the reactant which is entirely consumed in reaction.

The reaction is

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

1 mol 2 mol 1 mol

To find the limiting reagent, we consider the case of both reactants being completely consumed.

Case (I): Let CaCO₃ be completely consumed.

1 mol of CaCO₃ will give 1 mol of CaCl₂

Hence 10 mol of CaCO₃ will give 10 mol of CaCl₂.

Case (II): Let HCl be completely consumed.

2 mol of HCl will give 1 mol of CaCl₂

Hence, 0.19 mol of HCl will give 0.19/2 = 0.095 mol of CaCl₂.

Since the product formed in Case (II) is lesser, HCl is the limiting reagent, and the number of moles of CaCl₂ formed in the reaction is 0.095 mol. Mass of CaCl₂ formed will be 0.095 x Molar mass of CaCl₂ = 0.095 x 110 = 10.45g.