Question

Calcium carbons 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 gm of CaCO₃? Name the limiting reagent. Calculate the no. of moles of CaCl₂ formed in the reaction.

Answer 1

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

Number of moles of HCl = Volume in liters × Molarity

= \(\frac{250}{1000}\) x 0.76 = 0.19 mol.

Mass of CaCO₃ = 1000 g

Number of moles of CaCO₃

= \(\frac{Mass}{Molar\,mass}\) = \(\frac{1000\,g}{100\,g\,mol^{-1}}\)

1 mole of CaCO₃ reacts with 2 moles of HCI

10 moles of CaCO₃ will react with 2 × 10 = 20 moles of HCl

But, we have only 0.19 mol of HCl, therefore,

HCl is the limiting reactant.

2 moles of HCI give 1 mole of CaCl₂

0.19 mole of HCl will give = \(\frac{1}{2}\) x 0.19

= 0.095 moles of CaCl₂.