2. a) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
\(n_{H_2}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
\(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
b) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\)
=>\(m_{Al_2\left(SO_4\right)_3}=0,05.343=17,1\left(g\right)\)
a) \(n_{Al}=0,6\left(mol\right);n_{Fe_3O_4}=0,7\left(mol\right)\)
8Al + 3Fe3O4 ⟶ 4Al2O3 + 9Fe
Lập tỉ lệ : \(\dfrac{0,6}{8}< \dfrac{0,7}{3}\)
=> Sau phản ứng Fe3O4 dư
\(n_{Fe_3O_4\left(dư\right)}=0,7-\left(0,6.\dfrac{3}{8}\right)=0,475\left(mol\right)\)
=> \(m_{Fe_3O_4}=0,475.232=110,2\left(g\right)\)
b) \(n_{Al_2O_3}=0,3\left(mol\right);n_{Fe}=0,675\left(mol\right)\)
\(m_{Al_2O_3}=0,3.102=30,6\left(g\right)\)
\(m_{Fe}=0,675.56=37,8\left(g\right)\)
