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$\mathrm{CaCO}_3$ reacts with $\mathrm{HCl}$ to produce $\mathrm{CaCl}_2, \mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The approximate mass (in $\mathrm{g}$ ) of $\mathrm{CaCO}_3$
${ }_3$ required to react completely with $25 \mathrm{~mL}$ of $0.75 \mathrm{M} \mathrm{HCl}$ is (atomic mass of $\mathrm{Ca}=40, \mathrm{C}=12$, $\mathrm{O}=16, \mathrm{Cl}=35.5$ and $\mathrm{H}=1$ )
Options:
${ }_3$ required to react completely with $25 \mathrm{~mL}$ of $0.75 \mathrm{M} \mathrm{HCl}$ is (atomic mass of $\mathrm{Ca}=40, \mathrm{C}=12$, $\mathrm{O}=16, \mathrm{Cl}=35.5$ and $\mathrm{H}=1$ )
Solution:
2019 Upvotes
Verified Answer
The correct answer is:
0.94
$2 \mathrm{HCI}+\mathrm{CaCO}_3 \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
$25 \mathrm{~mL}$ of $0.75 \mathrm{~N}, \mathrm{HCl}=0.01875 \mathrm{~mole}$ of $\mathrm{HCl}$.
As per reaction twice the amount of $\mathrm{HCl}$ is needed in comparison to $\mathrm{CaCO}_3$.
Thus, amount of $\mathrm{CaCO}_3$ neutralised.
$$
\begin{aligned}
&=\frac{0.01875}{2}=0.009375 \mathrm{moI} \\
& 0.009375 \mathrm{moI} \times 100 \mathrm{~g} / \mathrm{moI}=0.9383 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \\
&=0.94 \mathrm{~g}
\end{aligned}
$$
$25 \mathrm{~mL}$ of $0.75 \mathrm{~N}, \mathrm{HCl}=0.01875 \mathrm{~mole}$ of $\mathrm{HCl}$.
As per reaction twice the amount of $\mathrm{HCl}$ is needed in comparison to $\mathrm{CaCO}_3$.
Thus, amount of $\mathrm{CaCO}_3$ neutralised.
$$
\begin{aligned}
&=\frac{0.01875}{2}=0.009375 \mathrm{moI} \\
& 0.009375 \mathrm{moI} \times 100 \mathrm{~g} / \mathrm{moI}=0.9383 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \\
&=0.94 \mathrm{~g}
\end{aligned}
$$
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