Ta có: 27nAl + 24nMg = 9 (1)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Mg}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{9}.100\%=60\%\\\%m_{Mg}=40\%\end{matrix}\right.\)
\(Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=9\\1,5a+b=0,45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,2.27}{9}.100\%=60\%;\%m_{Mg}=\dfrac{24.0,15}{9}.100\%=40\%\)
\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\\ n_{Al}=a,n_{Mg}=b\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=9\\1,5a+b=0,45\end{matrix}\right.\\ \Rightarrow a=0,2;b=0,15\\ \%m_{Al}=\dfrac{0,2.27}{9}\cdot100=60\%\\ \%m_{Mg}=100-60=40\%\)