Đáp án:
\(\begin{array}{l}
\% {m_{CuO}} = 66,67\% \\
\% {m_{MgO}} = 33,33\% \\
{V_{{\rm{dd}}HCl}} = 0,1l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
hh:MgO(a\,mol),CuO(b\,mol)\\
\left\{ \begin{array}{l}
40a + 80b = 12\\
95a + 135b = 23
\end{array} \right.\\
\Rightarrow a = b = 0,1\,mol\\
\% {m_{CuO}} = \dfrac{{0,1 \times 80}}{{12}} \times 100\% = 66,67\% \\
\% {m_{MgO}} = 100 - 66,67 = 33,33\% \\
{n_{HCl}} = 0,1 \times 2 + 0,1 \times 2 = 0,4\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,4}}{4} = 0,1l
\end{array}\)
