4Al + 3O2 $→$ 2Al2O3
$n_{O_2}=\dfrac{11,2}{22,4}=0,5(mol)$
Mình tính KL Al2O3
Theo PTHH $→n_{Al_2O_3}=\dfrac{2}{3}\times n_{O_2}=\dfrac{1}{3}(mol)$
$→m_{Al_2O_4}=\dfrac{1}{3}\times 102=34(g)$
C1: BTKL: mAl + mO2 = mAl2O3
$→$ mAl = 34 - 0,5 $\times 32=18(g)$
C2: Theo PTHH $→n_{Al}=\dfrac{4}{3}\times n_{O_2}=\dfrac{2}{3}(mol)$
$→m_{Al}=27\times \dfrac{2}{3}=18(g)$