Đáp án:

Phản ứng xảy ra:

\[ Mg+2HCl\to MgCl_2+H_2\qquad(1)\]

\[ CuO + 2HCl\to CuCl_2+H_2O\qquad(2)\]

\[n_{H_2}=\dfrac{0,224}{22,4}=0,01(mol)\]

Theo PTHH $(1)\to n_{Mg}=n_{H_2}=0,01(mol)$

$\to m_{Mg}=0,01.24=0,24(g)$
$\to \%m_{Mg}=\dfrac{0,24}{1,04}.100\%=23,07\%$

$\to \%m_{CuO}=100\%-23,07\%=76,93\%$

$n_{CuO}=\dfrac{1,04-0,24}{80}=0,01(mol)$

Theo PTHH $(1)\to n_{MgCl_2}=n_{Mg}=0,01(mol)$

Theo PTHH $(2)\to n_{CuO}=n_{CuCl_2}=0,01(mol)$
$\to m_{muoi}=0,01.95+0,01.135=2,3(g)$

PTPỨ

$
\begin{array}{l}
{\hspace{0.33em}{Mg}\mathrm{{+}}{2}{HCl}\mathrm{\longrightarrow}{MgCl}_{2}\mathrm{{+}}{H}_{2}\mathrm{\uparrow}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{CuO}\mathrm{{+}}{2}{HCl}\mathrm{\longrightarrow}{CuCl}_{2}\mathrm{{+}}{H}_{2}{O}}\\
{\hspace{0.33em}\hspace{0.33em}{nH}_{2}\mathrm{{=}}\frac{0\mathrm{,}\mathrm{242}}{\mathrm{22}\mathrm{,}4}\mathrm{{=}}{0}{\mathrm{,}}{0}{1}\hspace{0.33em}{\mathrm{(}}{mol}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}\mathrm{\Rightarrow}{nMg}\hspace{0.33em}\mathrm{{=}}\hspace{0.33em}{0}{\mathrm{,}}{0}{1}\hspace{0.33em}{\mathrm{(}}{mol}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}\mathrm{\Rightarrow}{mMg}\hspace{0.33em}\mathrm{{=}}\hspace{0.33em}{0}{\mathrm{,}}{0}{1}{\mathrm{.}}{2}{4}\hspace{0.33em}\mathrm{{=}}{0}{\mathrm{,}}{2}{4}{\mathrm{(}}{g}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}\mathrm{\Rightarrow}{\mathrm{\%}}{mMg}\mathrm{{=}}\frac{0\mathrm{,}\mathrm{24}}{1\mathrm{,}\mathrm{04}}\mathrm{\times}{1}{0}{0}{\mathrm{\%}}\mathrm{{=}}{2}{3}{\mathrm{,}}{0}{8}{\mathrm{\%}}}\\
{\hspace{0.33em}\hspace{0.33em}}\\
{\hspace{0.33em}\hspace{0.33em}\mathrm{\Rightarrow}{\mathrm{\%}}{mCuO}\mathrm{{=}}{\mathrm{100}}{\mathrm{\%}}\mathrm{{-}}{\mathrm{23}}{\mathrm{,}}{\mathrm{08}}{\mathrm{\%}}\mathrm{{=}}{\mathrm{76}}{\mathrm{,}}{\mathrm{92}}{\mathrm{\%}}}
\end{array}
$

$
\begin{array}{l}
{{mCuO}\mathrm{{=}}{1}{\mathrm{,}}{0}{4}\mathrm{{-}}{0}{\mathrm{,}}{2}{4}\mathrm{{=}}{0}{\mathrm{,}}{8}{\mathrm{(}}{g}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}{nCuO}\mathrm{{=}}\frac{0\mathrm{,}8}{\mathrm{80}}\mathrm{{=}}{0}{\mathrm{,}}{0}{1}{\mathrm{(}}{mol}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}{nHCl}\mathrm{{=}}{0}{\mathrm{,}}{0}{1}{\mathrm{.}}{2}\mathrm{{+}}{0}{\mathrm{,}}{0}{1}{\mathrm{.}}{2}\mathrm{{=}}{0}{\mathrm{,}}{0}{4}{\mathrm{(}}{mol}{\mathrm{)}}}
\end{array}
$

CM dung dịch HCl đã dùng là:

$
{C}_{M}\mathrm{{=}}\frac{n}{V}\mathrm{\times}{\mathrm{100}}{\mathrm{\%}}\mathrm{{=}}\frac{0\mathrm{,}\mathrm{04}}{0\mathrm{,}\mathrm{25}}\mathrm{{=}}{\mathrm{16}}{\mathrm{\%}}
$

Khối lượng muối thu được là:

$
\begin{array}{l}
{{MgCl}_{2}\mathrm{{=}}{0}{\mathrm{,}}{0}{1}{\mathrm{.}}{9}{5}\mathrm{{=}}{0}{\mathrm{,}}{9}{5}{\mathrm{(}}{g}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{mCuCl}_{2}\mathrm{{=}}{0}{\mathrm{,}}{0}{1}{\mathrm{.}}{1}{3}{5}\mathrm{{=}}{1}{\mathrm{,}}{3}{5}{\mathrm{(}}{g}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{M}\mathrm{{=}}{0}{\mathrm{,}}{\mathrm{95}}\mathrm{{+}}{1}{\mathrm{,}}{\mathrm{35}}\mathrm{{=}}{2}{\mathrm{,}}{3}{\mathrm{(}}{g}{\mathrm{)}}}
\end{array}
$