\(n_{Al}=\dfrac{3.5}{27}=\dfrac{7}{54}\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{180\cdot12.25\%}{98}=0.225\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2.........3\)
\(\dfrac{7}{54}.....0.225\)
\(LTL:\dfrac{\dfrac{7}{54}}{2}< \dfrac{0.225}{3}\Rightarrow H_2SO_4dư\)
\(n_{H_2}=\dfrac{7}{54}\cdot\dfrac{3}{2}=\dfrac{7}{36}\left(mol\right)\)
\(V_{H_2}=\dfrac{7}{36}\cdot22.4=4336\left(l\right)\)
\(m_{dd}=3.5+180-\dfrac{7}{36}\cdot2=183.11\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{7}{108}\cdot342}{183.11}\cdot100\%=12.1\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0.225-\dfrac{7}{36}\right)\cdot98}{183.11}\cdot100\%=1.63\%\)
