Chemistry Q74 – JEE-MAIN 2025

Home Chemistry Basic ChemistryQuestion #74

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

Enter Calculated Numerical Value:

Clue 1 of 3: Chemical Reaction

Identify the balanced chemical equation for the thermal decomposition of limestone.

🧠 Full Solution Path

Step 1: Write the balanced chemical equation and calculate molar masses✦ Active

Limestone is primarily calcium carbonate ( $C a C O_{3}$ ). Its thermal decomposition yields calcium oxide ( $C a O$ ) and carbon dioxide ( $C O_{2}$ ). The balanced chemical equation is:

C a C O_{3} (s) \to C a O (s) + C O_{2} (g)

Given molar masses: $Ca = 40$ , $O = 16$ , $C = 12$ .

Molar mass of $C a C O_{3} = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 g/mol$ .

Molar mass of $C a O = 40 + 16 = 56 g/mol$ .

Step 2: Calculate the mass of pure calcium carbonate○ Expand

The total mass of limestone is $150 kg$ , and it is $75 %$ pure. The actual mass of pure $C a C O_{3}$ available for decomposition is:

Mass of pure C a C O_{3} = 150 kg \times \frac{75}{100} = 112.5 kg

Step 3: Calculate the mass of calcium oxide produced○ Expand

From the balanced equation, $1 mole$ of $C a C O_{3}$ produces $1 mole$ of $C a O$ . This means $100 g$ of $C a C O_{3}$ produces $56 g$ of $C a O$ . We can use a stoichiometric ratio to find the mass of $C a O$ produced from $112.5 kg$ of $C a C O_{3}$ :

Mass of C a O = Mass of pure C a C O_{3} \times \frac{Molar mass of C a O}{Molar mass of C a C O_{3}}

Mass of C a O = 112.5 kg \times \frac{56 g/mol}{100 g/mol} = 112.5 \times 0.56 = 63 kg

The amount of calcium oxide produced is $63 kg$ . The nearest integer is $63$ .

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Chemistry Question 74 – JEE-MAIN 2025

Clue 1 of 3: Chemical Reaction

🧠 Full Solution Path