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$60 g m s$ of limestone on heating produced $22 g$ of ${C O}_{2}$ . The percentage of ${C a C O}_{3}$ in limestone is

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$80 %$

$60 %$

$83.3 %$

$87.66 %$

answer is C.

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Detailed Solution

Balanced chemical reaction

${C a C O}_{3} \to C a O + {C O}_{2}$

$22 g o f C O_{2} c o r r e s p o n d s t o 0.5 m o l e s .$

The molar mass of calcium carbonate is 100 g.

Mass of 0.5 moles of calcium carbonate is 50g.

The percentage of ${C a C O}_{3}$ in limestone is $\frac{50}{60} \times 100 = 83.3 %$ .

Therefore, option C is correct.

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