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When 10 g of 90% pure limestone is heated,the approximate volume (in L) of ${C O}_{2}$ liberated at STP is

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4.4

2.0

4.0

22.4

answer is B.

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Detailed Solution

${C a C O}_{3} \overset{∆}{⟶} C a O + {C O}_{2}$

Ratio of moles is 1:1.

10 g of 90% pure limestone = $9 g {C a C O}_{3}$

Molar Weight of ${C a C O}_{3} = 100 g / m o l$

$9/100 moles {CO}_{2} gives x liters at STP.

$x = 9 / 100 \times 22.4 = 2.016 L$

Hence the correct option is (B) 2.0.

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