The mass of $C O_{2}$ obtained by the complete thermal decomposition of $80 %$ of $40$ g pure sample of $C a C O_{3}$ is nearly equal to?

4.4

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44

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32

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14

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Solution

The correct option is D $14$ g
Weight of pure $C a C O_{3} = \frac{40}{100} \times 80 = 32$
Number of moles of $C a C O_{3} = \frac{32}{100} = 0.32$
$C a C O_{3} ⟶ C a O + C O_{2}$
$1$ mole of $C a C O_{3}$ gives $1$ mole of $C O_{2}$
$∴$ Moles of $C O_{2} = \frac{32}{100}$
$∴$ Weight of carbon dioxide= $\frac{32}{100} \times 44 = 14.08 g m$

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