4Al+3O2−to−>2Al2O3
2Mg+O2−to−>2MgO
nO2=\(\dfrac{33,6}{22,4}\)=1,5(mol)
nAl=\(\dfrac{2,7}{27}\)=0,1(mol)
Theo PT
nO2=\(\dfrac{3}{4}\)nAl=0,075(mol)
=>nO2(2)=1,5−0,075=1,425(mol)
nMg=2nO2=2,85(mol)
=>mMg=2,85.24=68,4(g)
=>mhỗnhợp=68,4+2,7=71,1(g)
=>%mAl=\(\dfrac{2.7}{71,1}\)100=3,8%
=>%mMg=100−3,8=96,2%