When 3.0 g of calcium carbonate ( $CaC O_{3}$ ) is heated strongly, it decomposes to calcium oxide ( $CaO$ ) and carbon dioxide ( $C O_{2}$ ). Calculate the volume of $C O_{2}$ gas produced at STP. (Molar masses: $CaC O_{3} = 100 g mol^{- 1}$ , $C O_{2} = 44 g mol^{- 1}$ ; $1 mol$ gas = $22.4 L$ at STP.)

medium 3 marks Class 10 · Science · Chemical Reactions and Equations

Answer

0.67 L

The correct answer is 0.67 L — Round to two significant figures (as the data given) → $0.67 L$ .

Step-by-step solution

Find moles of $CaC O_{3}$ : $n = \frac{3.0 g}{100 g mol ^{- 1}} = 0.030 mol$ .
The decomposition reaction $CaC O_{3} \to CaO + C O_{2}$ shows a 1:1 mole ratio between $CaC O_{3}$ and $C O_{2}$ , so $n_{C O_{2}} = 0.030 mol$ .
Convert moles of $C O_{2}$ to volume at STP: $V = n \times 22.4 L mol^{- 1} = 0.030 \times 22.4 = 0.672 L$ .
Round to two significant figures (as the data given) → $0.67 L$ .

Common mistake: Using the mass of

CaO

instead of

CaC O_{3}

to find moles.

Concept tested: Thermal decomposition stoichiometry