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The right option for the mass of CO2CO_2 produced by heating 20 g20\text{ g} of 20%20\% pure limestone is (Atomic mass of Ca=40Ca = 40) [CaCO3→1200 KCaO+CO2][CaCO_3 \xrightarrow{1200\text{ K}} CaO + CO_2]

A

1.12 g

B

1.76 g

C

2.64 g

D

1.32 g

Step-by-Step Solution

Mass of pure CaCO3=20 g×0.20=4 gCaCO_3 = 20\text{ g} \times 0.20 = 4\text{ g}. Molar mass of CaCO3=40+12+3(16)=100 g/molCaCO_3 = 40 + 12 + 3(16) = 100\text{ g/mol}. Moles of CaCO3=4/100=0.04 molCaCO_3 = 4/100 = 0.04\text{ mol}. From the reaction CaCO3→CaO+CO2CaCO_3 \rightarrow CaO + CO_2, 1 mole of CaCO3CaCO_3 produces 1 mole of CO2CO_2. So, 0.04 moles of CO2CO_2 are produced. Mass of CO2=0.04×44=1.76 gCO_2 = 0.04 \times 44 = 1.76\text{ g}.

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