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