\(n_{Mg}=a\left(mol\right)\)
\(n_{Al}=b\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)
Giả sử : hỗn hợp chỉ có Mg
\(n_{Mg}=\dfrac{7.8}{24}=0.325\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(0.325......0.325\)
\(n_{H_2SO_4}=0.325< 0.5\left(1\right)\)
Giả sử : hỗn hợp chỉ có Al.
\(n_{Al}=\dfrac{7.8}{27}=0.289\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.289....0.4335\)
\(n_{H_2SO_4}=0.4335\left(mol\right)< 0.5\left(2\right)\)
\(\left(1\right),\left(2\right):\)
Hỗn hợp tan hết , axit dư
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(\left\{{}\begin{matrix}24a+27b=7.8\\a+1.5b=0.4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)
\(\%Mg=\dfrac{0.1\cdot24}{7.8}\cdot100\%=30.77\%\)
\(\%Al=69.23\%\)
