20g of CaCO₃ reacts with HCl — find volume of CO₂ at STP

Question

20 g of calcium carbonate (CaCO₃) reacts completely with excess hydrochloric acid (HCl). Find the volume of CO₂ gas evolved at STP.

Solution — Step by Step

\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \uparrow

From the equation: 1 mole of CaCO₃ produces 1 mole of CO₂.

M(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \text{ g/mol}

n(\text{CaCO}_3) = \frac{\text{given mass}}{\text{molar mass}} = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ mol}

From the balanced equation, the mole ratio is:

n(\text{CO}_2) : n(\text{CaCO}_3) = 1 : 1

Therefore:

n(\text{CO}_2) = 0.2 \text{ mol}

At STP (Standard Temperature and Pressure: 0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 litres (molar volume).

V(\text{CO}_2) = n \times 22.4 = 0.2 \times 22.4 = \mathbf{4.48 \text{ L}}

Why This Works

The mole is the bridge between mass (measurable in the lab) and number of molecules or volume of gas (needed for stoichiometric calculations). The sequence is always:

\text{mass} \xrightarrow{\div M} \text{moles} \xrightarrow{\times \text{mole ratio}} \text{moles of product} \xrightarrow{\times 22.4} \text{volume at STP}

The molar volume (22.4 L/mol at STP) applies to all gases, regardless of the gas identity. This is a consequence of Avogadro’s law — equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Quick check: 20 g of CaCO₃ is exactly 0.2 mol (since $M = 100$ g/mol — a round number). This makes the arithmetic clean. In exams, CaCO₃ is a favourite because its molar mass is exactly 100 g/mol, making calculations straightforward.

Alternative Method

Using the formula directly: at STP, the volume of gas produced from mass $m$ of reactant with molar mass $M_R$ , where each mole of reactant gives $x$ moles of gas:

V = \frac{m \times x \times 22.4}{M_R} = \frac{20 \times 1 \times 22.4}{100} = 4.48 \text{ L}

This collapses all four steps into one calculation — useful for speed in MCQs.

Common Mistake

Students sometimes use 22.4 L/mol incorrectly for conditions other than STP, or confuse STP with NTP. STP = 0°C (273 K), 1 atm — molar volume 22.4 L/mol. NTP (Normal Temperature and Pressure) = 20°C (293 K), 1 atm — molar volume ≈ 24.04 L/mol. In Indian board exams (CBSE/JEE), unless stated otherwise, use 22.4 L/mol for STP problems.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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