To get $5.6$ lit of ${C O}_{2}$ at $S T P$ weight of ${C a C O}_{3}$ to be decomposed is

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$100 g$

$50 g$

$75 g$

$25 g$

answer is C.

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Detailed Solution

The reaction is ${C a C O}_{3} \to C a O + {C O}_{2}$ .

One mole of calcium carbonate forms one mole of carbon dioxide.

At STP, $5.6 L$ of carbon dioxide = $0.25$ moles.( since 22.4 L = 1 Mole )

They will be obtained from $0.25$ moles of calcium carbonate.

Molar mass of ${C a C O}_{3} = 100 g$ .

$0.25$ moles of calcium carbonate (molecular weight $100 g / m o l e$ ) corresponds to $25 g$ .

Hence option C is the correct answer.

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