Đáp án:
\(\begin{array}{l}
\% {m_{C{l_2}}} = 73,5\% \\
\% {m_{{O_2}}} = 26,5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Mg + {O_2} \to 2MgO\\
Mg + C{l_2} \to MgC{l_2}\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
hh\,X:C{l_2}(a\,mol),{O_2}(b\,mol)\\
BTKL:\\
{m_{hhX}} = 18,525 - 2,4 - 4,05 = 12,075g\\
\Rightarrow 71a + 32b = 12,075(1)\\
{n_{Al}} = \dfrac{{4,05}}{{27}} = 0,15\,mol\\
{n_{Mg}} = \dfrac{{2,4}}{{24}} = 0,1\,mol\\
BT\,e:2{n_{C{l_2}}} + 4{n_{{O_2}}} = 3{n_{Al}} + 2{n_{Mg}}\\
\Rightarrow 2a + 4b = 0,15 \times 3 + 0,1 \times 2(2)\\
(1),(2) \Rightarrow a = 0,125;b = 0,1\\
\% {m_{C{l_2}}} = \dfrac{{0,125 \times 71}}{{0,125 \times 71 + 0,1 \times 32}} \times 100\% = 73,5\% \\
\% {m_{{O_2}}} = 100 - 73,5 = 26,5\%
\end{array}\)
