a) 2Al+3H2SO4--->Al2(SO4)3+3H2
Al2O3+3H2SO4--->Al2(SO4)3+3H2O
Ta có
n\(_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Theo pthh1
n Al=\(\frac{2}{3}n_{H2}=0,1\left(mol\right)\)
m Al=\(0,1.27=2,7\left(g\right)\)
%m Al=\(\frac{2,7}{7,8}.100\%=34,6\%\)
%m Al2O3=100-34,6=65,4(g)
b)Theo pthh1
n H2SO4=n H2=0,15(mol)
Theo pthh2
n H2SO4=3n Al2O3=\(\frac{7,8-2,7}{102}.3=0,15\left(mol\right)\)
V H2SO4=(0,15+0,15)/2=0,15(l)
Lấy dư 10%
-->V H2SO4=\(0,15+\frac{0,15.10}{100}=0,165\left(l\right)\)