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NEET CHEMISTRYEasy

The mass of CO2CO_2 produced by heating 20 g of 20% pure limestone, as per the equation given below, is: CaCO3(s)→heatCaO(s)+CO2(g)CaCO_3(s) \xrightarrow{\text{heat}} CaO(s) + CO_2(g)

A

1.32 g

B

1.12 g

C

1.76 g

D

2.64 g

Step-by-Step Solution

  1. Calculate Mass of Pure Reactant:
  • Total mass of limestone sample = 20 g20 \text{ g}.
  • Percentage purity = 20%20\%.
  • Mass of pure CaCO3=20 g×20100=4 gCaCO_3 = 20 \text{ g} \times \frac{20}{100} = 4 \text{ g}.
  1. Determine Molar Masses:
  • Molar Mass of CaCO3(1×40+1×12+3×16)=100 g mol−1CaCO_3 (1 \times 40 + 1 \times 12 + 3 \times 16) = 100 \text{ g mol}^{-1} [Class 12 Chemistry, Ch 1, Example 1.2].
  • Molar Mass of CO2(1×12+2×16)=44 g mol−1CO_2 (1 \times 12 + 2 \times 16) = 44 \text{ g mol}^{-1}.
  1. Apply Stoichiometry:
  • According to the balanced equation, 1 mol CaCO31 \text{ mol } CaCO_3 yields 1 mol CO21 \text{ mol } CO_2.
  • 100 g CaCO3100 \text{ g } CaCO_3 yields 44 g CO244 \text{ g } CO_2.
  • Therefore, 4 g CaCO34 \text{ g } CaCO_3 yields: Mass of CO2=44100×4=1.76 g\text{Mass of } CO_2 = \frac{44}{100} \times 4 = 1.76 \text{ g}
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