Question

The right option for the mass of CO2 produced by heating 20g of 20% pure limestone is
(Atomic mass of Ca= 40)

• A. 1.76 g
• B. 2.64 g
• C. 1.32 g
• D. 1.12 g

Answer 1

Answer: (A)

1.76 g

Answer 2

Molecular mass of CaCO3 = 40+12+3x16 = 100 g/mol
The limestone is 20% pure, which means only 20% of it is calcium carbonate.
Moles of CaCO3 = $\frac {Mass\ of\ CaCO_3}{Molecular\ mass \ of\ CaCO_3 }$= $\frac {20 g \times 0.20}{100 \ g/mol}$ = 0.04 moles
Determine the moles of CO2 produced. From the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CO2. The molar ratio of CaCO3 to CO2 is 1:1.
Therefore, moles of CO2 produced = moles of CaCO3 = 0.04 moles
Molecular mass of CO2 = 12+2x16 = 44 g/mol
Mass of CO2 produced = (moles of CO2) x (molecular mass of CO2) = 0.04 moles x 44 g/mol = 1.76 g
So, the correct option is (A): 1.76 g