Calculate the mass of $C O_{2}$ produced by decomposing 40 g of 20 % pure limestone.

8 g

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4.48 g

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3.52 g

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17. 6 g

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Solution

The correct option is C 3.52 g
$C a C O_{3} \to C a O + C O_{2}$
Mass of 20% pure $C a C O_{3}$ = 20 % $\times$ 40 g = 8 g
Moles of $C a C O_{3} = \frac{given mass}{molar mass} = \frac{8}{100} = 0.08$ moles
As per the reaction,
1 mole of $C a C O_{3}$ decomposes to give 1 mole of $C O_{2}$ gas
0.08 mole of $C a C O_{3}$ will give 0.08 mole of $C O_{2} .$
mass of $C O_{2}$ produced = $0.08 \times 44$ = 3.52 g

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