a.Theo định luật bảo toàn khối lượng, ta có:
\(m_{Cl_2}=36,2-7,8=28,4g\)
\(n_{Cl_2}=\dfrac{m}{M}=\dfrac{28,4}{71}=0,4mol\)
\(2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\)
2 3 2 ( mol )
\(2Mg+Cl_2\rightarrow\left(t^o\right)2MgCl_2\)
2 1 2 ( mol )
Gọi \(n_{Al}=x\)
\(n_{Mg}=y\)
\(\Rightarrow m_{Al}=27x\)
\(\Rightarrow m_{Mg}=24y\)
Ta có:
\(\left\{{}\begin{matrix}m_{hh}=27x+24y=7,8\\\dfrac{3}{2}x+\dfrac{1}{2}y=0,4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{75}\\y=\dfrac{1}{25}\end{matrix}\right.\)
\(\Rightarrow m_{Al}=27.\dfrac{19}{75}=6,84g\)
\(\Rightarrow m_{Mg}=24.\dfrac{1}{25}=0,96g\)
\(\%m_{Al}=\dfrac{6,84.100}{7.8}=87,7\%\)
\(\%m_{Mg}=100\%-87,7\%=12,3\%\)
