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Law of Conservation of Mass

10.0 g of CaC...

Question

10.0 g of CaCO $_{3}$ on heating gave 4.4 g of CO $_{2}$ and 5.6 g of CaO. The observation is in agreement with the:

law of constant composition

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law of multiple proportions

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law of reciprocal proportion

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law of conservation of mass

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Solution

The correct option is D law of conservation of mass
$C a C O_{3} △ - \to C a O + C O_{2}$
$10.0 g$ $5.6 g$ $4.4 g$

Since, $M a s s o f r e a c t a n t = M a s s o f p r o d u c t$

$10.0 g = 4.4 g + 5.6 g$

$10.0 g = 10.0 g$

So, law of conservation of mass is constant.

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By heating 10g of $C a C O_{3}$ , 5.6g of CaO is formed. What is the moles of $C O_{2}$ obtained in this reaction?

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