4Al+3O2-to>2Al2O3
0,2---------\(\dfrac{2}{15}\) mol
n O2=\(\dfrac{6,4}{32}\)=0,2 mol
m Al2O3=\(\dfrac{2}{15}\).102=13,6g
\(n_{O_2}=\dfrac{m}{M}=\dfrac{6,4}{32}=0,2\left(mol\right)\)
\(4Al\) \(+\) \(3O_2\) → \(2Al_2O_3\)
\(0,2\) → \(\dfrac{2}{15}\) \(\left(mol\right)\)
\(m_{Al_2O_3}=n.M=\dfrac{2}{15}.102=13,6\left(g\right)\)