a,
mCl2= 26,7-5,4= 21,3g
=> nCl2= 21,3/71= 0,3 mol
2M+ 3Cl2 -> 2MCl3
=> nM= 0,2 mol
=> MM= 5,4/0,2= 27. Vậy M là nhôm (Al)
b,
2KMnO4+ 16HCl -> 2KCl+ 2MnCl2+ 5Cl2+ 8H2O
nCl2= 0,3 mol
=> nKMnO4= 0,12 mol; nHCl= 0,96 mol
mKMnO4= 0,12.158= 18,96g
mHCl= 0,96.36,5= 35,04g
=> mdd HCl= 35,04:36,5%= 96g
=> V dd HCl= 96/1,2= 80 ml