\(n_{Al}=0,3\left(mol\right);n_{H_2SO_4}=0,75\left(mol\right)\)
Dễ thấy \(H_2SO_4\) dư \(\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,15\left(mol\right);n_{H_2}=0,45\left(mol\right)\)
\(m_{\text{dd sau pư}}=100+8,1-0,45.2=107,2\left(g\right)\)
\(\Rightarrow C\%=\dfrac{0,15.342}{107,2}.100\%=47,85\%\)
Cái này đúng ra là \(C\%=47,85447761\%\approx47,86\%\)