nH2= 0,2(mol)
PHHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2
2/15________0,4_____2/16___0,2(mol)
mHCl= 0,4.36,5=14,6(g) -> mddHCl= (14,6.100)/15=292/3(g)
mAl= 2/15 . 27=3,6(g)
mAlCl3=133,5. 2/15=17,8(g)
mddA=mddAlCl3= mddHCl + mAl- mH2= 292/3 + 3,6 - 0,2.2=1508/15(g)
=> C%ddAlCl3= [17,8/(1508/15)].100=17,706%