4Al+3O2\(\rightarrow\)2Al2O3
a)
nAl=\(\frac{5,4}{27}\)=0,2(mol)
Theo pt ta có nAl2O3=\(\frac{nAl}{2}\)=\(\frac{0,2}{2}\)=0,1(mol)
\(\text{mAl2O3=0,1.102=10,2(g)}\)
b)
Theo pt nO2=\(\frac{3}{4}.nAl\)=\(\frac{3}{4}.0,2\)0,15(mol)
\(\text{VO2=0,15.22,4=3,36(l)}\)