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 \(0.75 \ M\) \(HCl\)?

Question

In the reaction, \[ 2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 6\text{Cl}^-(aq) + 3\text{H}_2(g) \]

Answer 1

Given:

Volume of HCl solution = 25 mL = 0.025 L
Molarity of HCl = 0.75 M
Molar mass of CaCO₃ = 100 g mol⁻¹

Step 1: Write the balanced chemical equation

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

Step 2: Calculate moles of HCl

Moles of HCl = Molarity × Volume

Moles of HCl = 0.75 × 0.025

Moles of HCl = 0.01875 mol

Step 3: Calculate moles of CaCO₃ required

From the balanced equation:
2 moles of HCl react with 1 mole of CaCO₃

Moles of CaCO₃ = 0.01875 / 2

Moles of CaCO₃ = 0.009375 mol

Step 4: Calculate mass of CaCO₃

Mass = Moles × Molar mass

Mass of CaCO₃ = 0.009375 × 100

Mass of CaCO₃ = 0.94 g

Final Answer:

Mass of calcium carbonate required to react completely with 25 mL of 0.75 M HCl is:
0.94 g

Solution and Explanation