JEE Main 2025 — Some Basic Concepts Of Chemistry Question with Solution

The reaction is:

${\mathrm{CaCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})\to {\mathrm{CaCl}}_2(\mathrm{aq})+{\mathrm{CO}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Calculate Moles of ${\mathrm{CaCO}}_3$:

The molar mass of ${\mathrm{CaCO}}_3$ is calculated as

$40+12+3\times 16=100g/mol$

Given 1000 g of ${\mathrm{CaCO}}_3$, the number of moles is

$\frac{1000}{100}=10\mathrm{moles}$

Calculate Moles of $\mathrm{HCl}$:

Using the given concentration and volume for $\mathrm{HCl}$, the moles are calculated as follows:

$0.76\times \frac{250}{1000}=0.19\mathrm{moles}$

$\mathrm{HCl}$ is the limiting reagent (L.R.) because it has fewer moles than needed to completely react with the ${\mathrm{CaCO}}_3$.

Calculate Moles of ${\mathrm{CaCl}}_2$ Formed:

According to the stoichiometry of the reaction, 2 moles of $\mathrm{HCl}$ produce 1 mole of ${\mathrm{CaCl}}_2$. Therefore, the moles of ${\mathrm{CaCl}}_2$ formed are:

$\frac{0.19}{2}=0.095\mathrm{moles}$

Calculate Mass of ${\mathrm{CaCl}}_2$:

The molar mass of ${\mathrm{CaCl}}_2$ is:

$40+2\times 35.5=111g/mol$

Thus, the mass of ${\mathrm{CaCl}}_2$ is:

$0.095\times 111=10.545\mathrm{g}$

Hence, the mass of ${\mathrm{CaCl}}_2$ formed is 10.545 g.