\(n_{H_2SO_4}=\dfrac{156.8\cdot15\%}{98}=0.24\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.16.......0.24..........0.08..........0.24\)
\(m_{dd}=0.16\cdot27+156.8-0.24\cdot2=160.64\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0.08\cdot342}{160.64}\cdot100\%=17.03\%\)
2Al + 3H2SO4 ⟶ Al2(SO4)3 + 3H2
\(n_{H_2SO_4}=\dfrac{15\%.156,8}{98}=0,24\left(mol\right)\)
Ta có :\(n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=0,16\left(mol\right)\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,08\left(mol\right)\)
\(n_{H_2}=n_{H_2SO_4}=0,24\left(mol\right)\)
\(m_{ddsaupu}=0,16.27+156,8-0,24.2=160,64\left(g\right)\)
=>\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,08.342}{160,64}.100=17,03\%\)
Cu + H2SO4 đặc, nóng