Molar mass of $\mathrm{CaCO}_3=40+12+3 \times 16=100 \mathrm{~g} \mathrm{~mol}^{-1}$

Moles of $\mathrm{CaCO}_3$ in $1000 \mathrm{~g}, n_{\mathrm{CaCO}_3}=\frac{\text { Mass }(\mathrm{g})}{\text { Molar mass }}$

$$\begin{aligned} &\begin{aligned} n_{\mathrm{CaCO}_3} & =\frac{1000 \mathrm{~g}}{100 \mathrm{~g} \mathrm{~mol}^{-1}}=10 \mathrm{~mol} \\ \text { Molarity } & =\frac{\text { Moles of solute }(\mathrm{HCl}) \times 1000}{\text { Volume of solution }} \end{aligned}\\ &\text { (It is given that moles of } \mathrm{HCl} \text { in } 250 \mathrm{~mL} \text { of } 0.76 \mathrm{M} \mathrm{HCl}=n_{\mathrm{HCI}} \text { ) }\\ &\begin{aligned} 0.76 & =\frac{n_{\mathrm{HCl}} \times 1000}{250} \\ n_{\mathrm{HCl}} & =\frac{0.76 \times 250}{1000}=0.19 \mathrm{~mol} . \end{aligned} \end{aligned}$$

$$\underset{1 \mathrm{~mol}}{\mathrm{CaCO}_3(\mathrm{~s})}+\underset{2 \mathrm{~mol}}{2 \mathrm{HCl}(\mathrm{aq})} \longrightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)$$

According to the equation,

1 mole of $\mathrm{CaCO}_3$ reacts with 2 moles HCl

$\therefore \quad 10$ moles of $\mathrm{CaCO}_3$ will react with $\frac{10 \times 2}{1}=20$ moles HCl .

But we have only 0.19 moles HCl , so HCl is limiting reagent and it limits the yield of $\mathrm{CaCl}_2$.

Since, $\quad 2$ moles of HCl produces 1 mole of $\mathrm{CaCl}_2$

$$\begin{aligned} 0.19 \text { mole of } \mathrm{HCl} \text { will produce } \frac{1 \times 0.19}{2} & =0.095 \mathrm{~mol} \mathrm{CaCl}_2 \\ \text { Molar mass of } \mathrm{CaCl}_2 & =40+(2 \times 35.5)=111 \mathrm{~g} \mathrm{~mol}^{-1} \\ \therefore \quad 0.095 \text { mole of } \mathrm{CaCl}_2 & =0.095 \times 111=10.54 \mathrm{~g} \end{aligned}$$

'Law of multiple proportions' was first studied by Dalton in 1803 which may be defined as follows

When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another.

e.g., hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.

$$\begin{aligned} &\begin{aligned} & \underset{2 \mathrm{~g}}{\text { Hydrogen }}+\underset{32 \mathrm{~g}}{\text { Oxygen }} \longrightarrow \underset{34 \mathrm{~g}}{\text { Hydrogen peroxide }} \end{aligned}\\ &\underset{2 \mathrm{~g}}{\text { Hydrogen }}+\underset{32 \mathrm{~g}}{\text { Oxygen }} \longrightarrow \underset{34 \mathrm{~g}}{\text { Hydrogen peroxide }} \end{aligned}$$

Here, the masses of oxygen (i.e., 16 g and 32 g ) which combine with a fixed mass of hydrogen $(2 \mathrm{~g})$ bear a simple ratio, i.e., $16: 32$ or $1: 2$.

As we know that, when compounds mixed in different proportionation, Then they form different compounds. In the above examples, when hydrogen is mixed with different proportion of oxygen, then they form water or hydrogen peroxide.

It shows that there are constituents which combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine into molecules.

Mass of $B$ which is combined with fixed mass of $A$ (say 1 g ) will be $2.5 \mathrm{~g}, 5 \mathrm{~g}, 1.25 \mathrm{~g}$ and 3.75 g . They are in the ratio $2: 4: 1: 3$ which is a simple whole number ratio. Hence, the law of multiple proportions is applicable.