JEE Main 2025 Apr 4 Shift 2, Chemistry Q22: Introduction to Metallurgical Processes

JEE Main2025Apr 4, Shift 2Chemistry

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is ______ kg. (Nearest integer)

Given : Molar mass (in g mol $^{- 1}$ ) of Ca-40, O-16, C-12

Your answer

Solution

Decomposition: $C a C O_{3} Δ C a O + C O_{2}$ .

Pure $C a C O_{3}$ in 150 kg of 75% pure limestone:

$m (C a C O_{3}) = 150 \times 0.75 = 112.5$ kg.

Moles of $C a C O_{3} = \frac{112500}{100} = 1125$ mol (using $M (C a C O_{3}) = 100$ g mol $^{- 1}$ ).

From stoichiometry, moles of $C a O$ produced = 1125 mol. With $M (C a O) = 56$ g mol $^{- 1}$ :

$m (C a O) = \frac{1125 \times 56}{1000} = 63$ kg.

$∴$ The amount of CaO produced is 63 kg.

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