What quantity of lime stone on heating will give $56 k g$ of $C a O$ ?

1000 k g

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56 k g

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44 k g

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100 k g

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Solution

The correct option is C $100 k g$
$C a C O_{3} Δ - \to C a O + C O_{2} ↑$

Molar masses of $C a O$ and $C a C O_{3}$ are 56 g/mol and 100 g/mol respectively.

Mass of $C a O = 56 k g = 56000 g$

$(∵ 1 k g = 1000 g)$

Number of moles of $C a O = \frac{56000 g}{56 g / m o l} = 1000 m o l$

As per balanced chemical equation, 1 mole of $C a O$ is obtained by heating 1 mole of $C a C O_{3}$ .

Hence, 1000 moles of $C a O$ will be obtained by heating 1000 moles of $C a C O_{3}$ .

Mass of $C a C O_{3} = 1000 m o l \times 100 g / m o l = 10^{5} g$

Convert unit from g to kg.

Mass of $C a C O_{3} = 10^{5} g \times \frac{1 k g}{1000 g} = 100 k g$

Hence, the correct option is $D$

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