Đặt số mol

\(n_{O_2^{}}^{} = a\,{\text{mol; n}}_{Cl_2^{}}^{} = b\;{\text{mol}}\)

\({\text{n}}_{hh}^{} = \frac{{7,84}}{{22,4}} = 0,35\;{\text{mol}}\)

\({\text{BTKL:m}}_{hh}^{} + m_{KL}^{} = m_Z^{}\)

\( \to m_{hh}^{} = 30,1 - 11,1 = 19g\)

\(32a + 71b = 19\)

\(a + b = 0,35\)

\( \to a = 0,15;b = 0,2\)

\(n_{Al}^{} = x{\text{ mol;n}}_{Mg}^{} = y{\text{ mol}}\)

\(4Al + 3O_2^{}\xrightarrow{{{t^o}}}2Al_2^{}O_3^{}\)

\(2Al + 3Cl_2^{}\xrightarrow{{{t^o}}}2AlCl_3^{}\)

\(2Mg + O_2^{}\xrightarrow{{{t^o}}}2MgO\)

\(Mg + Cl_2^{}\xrightarrow{{{t^o}}}MgCl_2^{}\)

\(n_{Al}^{} = \frac{4}{3}n_{O_{2_{(1)}^{}}^{}}^{} + \frac{2}{3}n_{Cl_{2_{(2)}^{}}^{}}^{}\)

\( \to 3n_{Al}^{} = 4n_{O_{2_{(1)}^{}}^{}}^{} + 2n_{Cl_{2_{(2)}^{}}^{}}^{}( + )\)

\(n_{Mg}^{} = 2n_{O_{2_{(3)}^{}}^{}}^{} + 1n_{Cl_{2_{(4)}^{}}^{}}^{}\)

\( \to 2n_{Mg}^{} = 4n_{O_{2_{(3)}^{}}^{}}^{} + 2n_{Cl_{2_{(4)}^{}}^{}}^{}( +  + )\)

Cộng (+) theo (++)

\(( + ),( +  + ) \to 3n_{Al}^{} + 2n_{Mg}^{} = 4n_{O_{2_{(1)}^{}}^{}}^{} + 2n_{Cl_{2_{(2)}^{}}^{}}^{} + 4n_{O_{2_{(3)}^{}}^{}}^{} + 2n_{Cl_{2_{(4)}^{}}^{}}^{}\)

\( \to 3n_{Al}^{} + 2n_{Mg}^{} = 4(n_{O_{2_{(1)}^{}}^{}}^{} + n_{O_{2_{(3)}^{}}^{}}^{}) + 2(n_{Cl_{2_{(2)}^{}}^{}}^{} + n_{Cl_{2_{(4)}^{}}^{}}^{})\)

\( \to 3n_{Al}^{} + 2n_{Mg}^{} = 4n_{O_{2_{PU}^{}}^{}}^{} + 2n_{Cl_{2_{PU}^{}}^{}}^{}\)

\( \to 3n_{Al}^{} + 2n_{Mg}^{} = 4.0,15 + 2.0,2\)

\(27x + 24y = 11,1\)

\( \to x = 0,1;y = 0,35\)

\(\% m_{Al}^{} = \frac{{0,1.27}}{{11,1}}.100 = 24,32\% \)