Question

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction,

\[\ce{CaCO_{3(s)} + 2 HCl_{(aq)} → CaCl_{2(aq)} + CO_{2(g)} + H2O_{(l)}}\]

What mass of CaCO₃ 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 1 L of water

≡ [(0.75 mol) × (36.5 g mol^-1)] HCl is present in 1 L of water

≡ 27.375 g of HCl is present in 1 L of water

Thus, 1000 mL of solution contains 27.375 g of HCl.

∴ Amount of HCl present in 25 mL of solution

`=(27.375  "g")/(1000  "mL")xx25  "mL"`

= 0.6844 g

From the given chemical equation,

\[\ce{CaCO_{3(s)} + 2 HCl_{(aq)} → CaCl_{2(aq)} + CO_{2(g)} + H2O_{(l)}}\]

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO₃ (100 g).

∴ Amount of CaCO₃ that will react with 0.6844 g `=100/73xx0.6844`

= 0.9375 g