$10.0 g$ of Calcium carbonate on heating gave $4.4 g$ of Calcium oxide and $5.6 g$ of Carbon dioxide. Show that these observations are in agreement with law of conservation of mass.

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Solution

Step 1: Law of conservation of mass:
The law of conservation of mass states that mass can neither be created nor be lost but it can be conserved.
Thus, from the law of conservation of mass, the total mass of reactants is equal to the total mass of products in any chemical reaction.
Step 2: Identify reactants and products:
From the given data,
$10.0 g$ of Calcium carbonate or limestone $({CaCO}_{3})$ on heating gave $4.4 g$ of Calcium oxide or Quick lime $(CaO)$ and $5.6 g$ of Carbon dioxide $({CO}_{2})$ .
${CaCO}_{3} (s) \overset{∆}{\to} CaO (s) + {CO}_{2} (g)$
Thus,
The reactant is $10.0 g$ of Calcium carbonate.
Products are $4.4 g$ of Calcium oxide and $5.6 g$ of Carbon dioxide.
Step 3: Apply the law of conservation of mass:
According to the law of conservation of mass,
Total mass of reactants $=$ Total mass of products
Thus, the total mass of Calcium carbonate should be equal to the total mass of Calcium oxide and Carbon dioxide together.
$\Rightarrow 10 g = 4.4 g + 5.6 g$
Hence, it is proved that the results are in agreement with the law of conservation of mass.

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10.0g of Calcium carbonate on heating gave 4.4g of Calcium oxide and 5.6g of Carbon dioxide. Show that these observations are in agreement with law of conservation of mass.

$10.0 g$ of Calcium carbonate on heating gave $4.4 g$ of Calcium oxide and $5.6 g$ of Carbon dioxide. Show that these observations are in agreement with law of conservation of mass.