PTHH: $4Al+3O_{2}\rightarrow 2Al_{2}O_{3}$
Ta có: $n_{Al}=\frac{8,1}{27}=0,3(mol)$
Theo pt: $n_{O_{2}}=\frac{3}{4}.n_{Al}=\frac{3}{4}.0,3=0,225(mol)$
$\Rightarrow V_{O_{2}}=n.22,4=0,225.22,4=5,04(l)$
$\Rightarrow V_{KK}=\frac{V_{O_{2}}.100}{20}=\frac{5,04.100}{20}=25,2(l)$
Theo pt...