Đáp án:
a) PTHH : `2Al+6HCl→2AlCl_3+3H_2↑`
`m_(HCl)=frac{5,475.200}{100}=10,95g`
`n_(HCl)=frac{10,95}{36,5}=0,3`mol
Theo pthh ta có : `n_(Al)=0,1`mol
`⇒m_(Al)=0,1.27=2,7g`
b) Theo pthh ta có : `n_(AlCl_3)=0,1`mol ; `n_(H_2)=0,15`mol
`⇒m_(AlCl_3)=0,1.133,5=13,35g`
`⇒V_(H_2)=0,15.22,4=3,36` lít (đktc)
c) `mdd=mAl+mdd_HCl-mH_2=2,7+200-0,15=202,55g`
`⇒C%_(AlCl_3)=frac{13,35}{202,55}100≈6,6%`
$\text{Shield Knight}$