Question

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O(l) What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Answer 1

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

0.75 M solution of HCl = 0.75 moles of HCl in 1 L water

Number of moles of HCl in 25mL of 0.75 M solution = (0.75/1000) X 25 = 0.019 moles

From this equation,

it is clear that one mole of CaCO3 reacts with two moles of HCl

So,

Number of moles of CaCO3 needed to react with 0.019 moles of HCl = (1/2) X 0.019 = 0.001

Mass of 0.001 mole CaCO3 = 0.001 X Molar mass of CaCO3 = 0.001 X 100 = 0.1g