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$15 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with
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Verified Answer
The correct answer is:
10.95 g of HCl
The balanced equation the reaction is
$\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3}+\underset{73 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
$100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with $73 \mathrm{~g}$ of $\mathrm{HCl}$
$\begin{aligned}
\text { So, } 15 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text { will react with } & =\frac{73}{100} \times 15 \mathrm{~g} \text { of } \mathrm{HCl} \\
& =10.95 \mathrm{~g} \text { of } \mathrm{HCl}
\end{aligned}$
$\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3}+\underset{73 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
$100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ completely reacts with $73 \mathrm{~g}$ of $\mathrm{HCl}$
$\begin{aligned}
\text { So, } 15 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text { will react with } & =\frac{73}{100} \times 15 \mathrm{~g} \text { of } \mathrm{HCl} \\
& =10.95 \mathrm{~g} \text { of } \mathrm{HCl}
\end{aligned}$
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