Question

2 gram of CaCO3 on strong heating gave 1.12 gram of CaO and 448 ml of CO2 show that the results illustrate the law of conservation of mass.

Answer 1

$\begin{minipage}{10cm}\underline{\large\textbf{Law of Conservation of Mass}}\begin{flushright}\small\text{Proposed by Antoine Lavoisier}\end{flushright}\\\\\begin{justified}``\textit{In a chemical reaction, total mass of products obtained is equal to that of reactants. Mass is neither created nor destroyed during a chemical reaction.}"\end{justified}\end{minipage}$

According to the question, $\sf{2\ g}$ of $\sf{CaCO_3}$ is decomposed to form $\sf{1.12\ g}$ of $\sf{CaO}$ and $\sf{448\ mL=0.448\ L}$ of $\sf{CO_2}$ at STP.

Mass of $\sf{22.4\ L}$ of $\sf{CO_2}$ is $\sf{44\ g.}$ Hence mass of $\sf{0.448\ L}$ of

$\sf{44\times\dfrac{0.448}{22.4}=0.88\ g}$

So the total mass of products is $\sf{1.12\ g+0.88\ g=2\ g.}$ This is the same as mass of reactant, or $\sf{CaCO_3.}$

Hence the result illustrates Law of Conservation of Mass.