Đáp án:
\(\begin{array}{l}
1)\\
\% {m_{A{l_2}{O_3}}} = 38,93\% \\
\% {m_{MgO}} = 61,07\% \\
2)\\
{C_M}AlC{l_3} = 0,2M\\
{C_M}MgC{l_2} = 0,4M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{HCl}} = 0,5 \times 1,4 = 0,7\,mol\\
hh:A{l_2}{O_3}(a\,mol),MgO(b\,mol)\\
\left\{ \begin{array}{l}
102a + 40b = 13,1\\
6a + 2b = 0,7
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,2\\
\% {m_{A{l_2}{O_3}}} = \dfrac{{0,05 \times 102}}{{13,1}} \times 100\% = 38,93\% \\
\% {m_{MgO}} = 100 - 38,93 = 61,07\% \\
2)\\
{n_{AlC{l_3}}} = 2{n_{A{l_2}{O_3}}} = 0,1\,mol\\
{n_{MgC{l_2}}} = {n_{MgO}} = 0,2\,mol\\
{C_M}AlC{l_3} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
{C_M}MgC{l_2} = \dfrac{{0,2}}{{0,5}} = 0,4M
\end{array}\)
