A sample of limestone is $20 %$ pure. Find the mass of $C O_{2}$ produced by decomposing $40 g$ of that sample of limestone.

7.04 g

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1.76 g

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3.52 g

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3 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 $C a C O_{3} = \frac{20}{100} \times 40 g = 8 g$
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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