\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PTHH :
\(4Al+3O_2\rightarrow2Al_2O_3\)
2/15 0,1 1/15
\(\dfrac{0,1}{3}< \dfrac{0,1}{2}\)
---> Tính theo O2
\(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)