Đáp án:
\(\begin{array}{l}
\% {V_{C{l_2}}} = 33,33\% \\
\% {V_{{O_2}}} = 66,67\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Mg + {O_2} \to 2MgO\\
Mg + C{l_2} \to MgC{l_2}\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
b)\\
{m_{hhX}} = 23,9 - 10,4 = 13,5g\\
{M_{hhX}} = 22,5 \times 2 = 45g/mol\\
\Rightarrow {n_{hh}} = \dfrac{{13,5}}{{45}} = 0,3\,mol\\
hhX:C{l_2}(a\,mol),{O_2}(b\,mol)\\
a + b = 0,3\\
71a + 32b = 13,5\\
\Rightarrow a = 0,1;b = 0,2\\
\% {V_{C{l_2}}} = \dfrac{{0,1}}{{0,1 + 0,2}} \times 100\% = 33,33\% \\
\% {V_{{O_2}}} = 100 - 33,33 = 66,67\%
\end{array}\)
