Đáp án:
`4Al+3O_2->2Al_2O_3`(1)
`2Mg+O_2->2MgO`(2)
`nMg=frac{4,8}{24}=0,2`mol
`nO_2=frac{15,68}{22,4}=0,7`mol
Theo ptpu ta có : `nO_2(1)=0,1`mol
Theo ptpu ta có : `nAl=frac{4}{3}nO_2(2)=0,8` mol
`mAl=0,8.27=21,6`g
`=>m_(hh)=26,4`g
`%mAl=frac{21,6}{26,4}.100=81,8`%
`%mMg=frac{4,8}{26,4}.100=18,2`%
$\text{Shield Knight}$