Question

Calcium carbonate reacts with aqueous \(HCl\) to give \(CaCl_2 \) and \(CO_2\) according to the reaction, \(CaCO_3 (s) + 2 HCl (aq) → CaCl_2 (aq) + CO_2(g) + H_2O(l)\). What mass of \(CaCO_3\) is required to react completely with \(25 \ mL\) of \(0.75 \ M\) \(HCl\)?

Answer 1

0.75 M of HCl ≡ 0.75 mol of HCl are present in 1L of water
≡ [(0.75 mol) × (36.5 g mol^-1)] HCl is present in 1L of water
≡ 27.375 g of HCl is present in 1L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
∴ Amount of HCl present in 25 mL of solution = \(\frac {27.375\ g }{1000\ mL} × 25 \ mL\) 
= 0.6844 g
From the given chemical equation, 
\(CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) + H_2O(l)\)
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO₃ (100 g).
Amount of CaCO₃ that will react with 0.6844 g = \(\frac {100}{71} × 0.6844\  g\)
= 0.9639 g