Calculate the mass of $C O_{2}$ produced by decomposing 40 g of 60 % pure limestone.

10.56 g

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12.38 g

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8.20 g

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14.52 g

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Solution

The correct option is A 10.56 g
$C a C O_{3} \to C a O + C O_{2}$
Mass of 60% pure $C a C O_{3}$ = 60 % $\times$ 40 g = 24 g
Moles of $C a C O_{3} = \frac{given mass}{molar mass} = \frac{24}{100} = 0.24$ moles
As per the reaction,
1 mole of $C a C O_{3}$ decomposes to give 1 mole of $C O_{2}$ gas
0.24 moles of $C a C O_{3}$ will give 0.24 mole of $C O_{2} .$
mass of $C O_{2}$ produced = $0.24 \times 44$ = 10.56 g

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