\(nAl=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(nO_2=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)
PTHH:
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
4 3 2 (mol)
0,3 0,225 0,15 (mol)
LTL :
\(\dfrac{0,3}{4}=\dfrac{0,225}{3}\)
=> Al đủ , O2 đủ
=> \(mAl_2O_3=0,15.102=15,3\left(g\right)\)