The volume of $C O_{2}$ (in litres) liberated at STP when 10 g of 90 % pure limestone is heated completely, is:

22.4 L

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2.24 L

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20.16 L

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2.016 L

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Solution

The correct option is D 2.016 L
$C a C O_{3} \to C a O + C O_{2}$
1 mole $C a C O_{3}$ produces 1 mole $C O_{2}$ .

We have $\frac{90}{100} \times 10$ = 9 g pure $C a C O_{3}$

Molar mass $C a C O_{3}$ = 100 g/mole

9 g $C a C O_{3}$ = $\frac{9}{100}$ = 0.09 mole $C a C O_{3}$ .

This will produce 0.09 mole $C O_{2}$ .
At S.T.P, $1$ mole $= 22.4$ L
$0.09$ mole $= 2.016$ L

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