Question 50: The right option for the mass of \mathrm{CO}_{2} produced by heating 20 g of 20 \% pure limestone is (Atomic mass of \mathrm{Ca}=40 )
\mathrm{CaCO}_{3} \xrightarrow{1200 \mathrm{~K}} \mathrm{CaO}+\mathrm{CO}_{2}(1) 2.64 g
(2) 1.32 g
(3) 1.12 g
(4) 1.76 g
Answer: Option (4)
Explanation:
Limestone mainly contains calcium carbonate, \mathrm{CaCO}_{3}.
Given that the limestone is only 20 \% pure,
the actual mass of \mathrm{CaCO}_{3} present in 20 g limestone is:
\text{Mass of } \mathrm{CaCO}_{3} = 20 \times \frac{20}{100} = 4 \text{ g}Molar mass of \mathrm{CaCO}_{3} is:
40 + 12 + 3 \times 16 = 100 \text{ g mol}^{-1}According to the reaction, 100 g of \mathrm{CaCO}_{3} produces 44 g of \mathrm{CO}_{2}.
Therefore, 4 g of \mathrm{CaCO}_{3} will produce:
\mathrm{CO}_{2} = \frac{44}{100} \times 4 = 1.76 \text{ g}Hence, the correct option is Option (4).