NEET 2023, Chemistry Q26: Stoichiometry

NEET2023Chemistry

The right option for the mass of CO $_{2}$ produced by heating $20$ g of $20%$ pure limestone is (Atomic mass of Ca $= 40$ )

$CaCO_{3} 1200 K CaO + CO_{2}$

Solution

Mass of pure $CaCO_{3} = 20% \times 20 = 4$ g.

Moles of $CaCO_{3} = \frac{4}{100} = 0.04$ mol.

By the 1:1 stoichiometry, $CO_{2}$ produced is also $0.04$ mol.

Mass of $CO_{2} = 0.04 \times 44 = 1.76$ g.

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