Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
a, \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
b, \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,075\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,075.24,79=1,85925\left(l\right)\)