\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Ta có :
\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow n_{Fe}=n_{H2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Al2O3}=94,2-0,15.56=85,8\left(g\right)\)
\(n_{Al2O3}=\frac{58,8}{102}=0,84\left(g\right)\)
\(\Rightarrow n_{H2SO4}=0,15+0,84.3=2,67\left(mol\right)\)
\(\Rightarrow V_{H2SO4}=\frac{2,67}{2}=1,335\left(l\right)\)
\(Fe+H2SO4-->H2SO4+H2\)
\(Al2O3+3H2SO4-->Al2\left(SO4\right)3+3H2O\)
\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Fe}=n_{H2}=0,15\left(mol\right)\)
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(m_{Al2O3}=94,2-8,4=85,8\left(g\right)\)
\(n_{H2SO4}=n_{H2}=0,15\left(mol\right)\)
\(n_{Al2O3}=\frac{85,8}{102}\approx0,84\left(mol\right)\)
\(n_{H2SO4}=3n_{Al2O3}=2,52\left(mol\right)\)
\(\sum n_{H2SO4}=0,15+2,52=2,67\left(mol\right)\)
\(V_{H2SO4}=\frac{2,67}{2}=1,335\left(l\right)\)