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$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}$ was added to $1 \mathrm{~g}$ of pure $\mathrm{CaCO}_{3}$. What would remain after the reaction?
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Verified Answer
The correct answer is:
Neither $\mathrm{CaCO}_{3}$ nor $\mathrm{HCl}$
Number of moles of $\mathrm{CaCO}_{3}=\frac{W}{M}=\frac{1}{100}$ $=0.01 \mathrm{~mol}$
$\therefore$ Molar mass of $\mathrm{CaCO}_{3}=100 \mathrm{~g} \mathrm{~mol}^{-1}$
$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}=\frac{100}{1000} \times \frac{1}{5}=0.02 \mathrm{~mol}$
The reaction is
$\therefore 0.01$ mole of $\mathrm{CaCO}_{3}$ reacts with $0.02$ mole of $\mathrm{HCl}$.
$\therefore$ Both are completely consumed in the given reaction.
$\therefore$ Molar mass of $\mathrm{CaCO}_{3}=100 \mathrm{~g} \mathrm{~mol}^{-1}$
$100 \mathrm{~mL}$ of $\frac{N}{5} \mathrm{HCl}=\frac{100}{1000} \times \frac{1}{5}=0.02 \mathrm{~mol}$
The reaction is
$\therefore 0.01$ mole of $\mathrm{CaCO}_{3}$ reacts with $0.02$ mole of $\mathrm{HCl}$.
$\therefore$ Both are completely consumed in the given reaction.
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