A student weighs out 12.0 g of calcium carbonate (CaCO3) and reacts it with an excess of dilute hydrochloric acid. The balanced equation is CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). What is the maximum volume of carbon dioxide gas produced at 25°C and 1.00 atm? (Molar mass of CaCO3 = 100.1 g/mol; R = 0.0821 L·atm·mol⁻¹·K⁻¹)

Correct answer

A. 2.94 L

Principle or equation

Amount-of-substance calculations relate mass to moles via molar mass, and stoichiometry gives the moles of product. The ideal gas law (PV = nRT) converts moles of gas to volume under given conditions.

Why this answer is correct

Moles of CaCO3 = 12.0 g / 100.1 g/mol = 0.1199 mol. From the balanced equation, 1 mol CaCO3 produces 1 mol CO2, so n(CO2) = 0.1199 mol. Using PV = nRT, V = nRT/P = (0.1199 mol)(0.0821 L·atm·mol⁻¹·K⁻¹)(298 K) / (1.00 atm) ≈ 2.94 L.

Example

For 10.0 g of CaCO3 (0.0999 mol), the CO2 volume at STP (273 K, 1 atm) would be 0.0999 × 0.0821 × 273 ≈ 2.24 L.