Calcium carbonate reacts with hydrochloric acid as: $CaC O_{3} + 2HCl \to CaC l_{2} + C O_{2} + H_{2} O$ . If $11.0 g$ of $CaC O_{3}$ is completely reacted with excess $HCl$ , calculate the volume of carbon dioxide gas produced at NTP (1 mol gas = $22.4 L$ ).

easy 2 marks Class 10 · Science · Carbon and its Compounds

Answer

2.46 L of

C O_{2}

at NTP

The correct answer is 2.46 L of $C O_{2}$ at NTP — State the final volume.

Step-by-step solution

Calculate the molar mass of $CaC O_{3}$ : $40.08 + 12.01 + 3 \times 16.00 = 100.09 g mol^{- 1}$ .
Convert the given mass to moles: $n = \frac{11.0 g}{100.09 g mol ^{- 1}} = 0.1098 mol$ .
From the balanced equation, 1 mol $CaC O_{3}$ produces 1 mol $C O_{2}$ , so moles of $C O_{2}$ formed = 0.1098 mol.
Use the gas volume relation at NTP: $V = n \times 22.4 L mol^{- 1} = 0.1098 \times 22.4 = 2.46 L$ .
State the final volume.

Common mistake: Forgetting to convert grams of

CaC O_{3}

to moles before using the gas volume relationship.

Concept tested: Mole concept and gas volume