Question

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2HCl (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

1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g 

∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/1000 × 25 g = 0.6844 g 

From the given chemical equation, 

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

2 mol of HCl i.e. 73g HCl react completely with 1 mol of CaCO₃ i.e. 100g

∴ 0.6844 g HCl reacts completely with CaCO₃ = 100/73 × 0.6844 g = 0.938 g