a) PTHH : \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\).
b) Ta có : \(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{20,4}{102}=0,2\left(mol\right)\).
Theo phương trình, \(n_{Al}=2n_{Al_2O_3}=2.\left(0,2\right)=0,4\left(mol\right)\).
\(\Rightarrow m_{Al}=n_{Al}.M_{Al}=\left(0,4\right).27=10,8\left(g\right)\)
c) Theo phương trình và ý b) : \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=\dfrac{3}{2}\cdot0,2=0,3\left(mol\right)\).
\(\Rightarrow V_{O_2}=n_{O_2}.22,4=\left(0,3\right).\left(22,4\right)=6,72\left(l\right)\).
\(\Rightarrow V_{KK}=5V_{O_2}=5.\left(6,72\right)=33,6\left(l\right)\).