a/ PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
b/ Ta có: nH2= 6,72: 22,4= 0,3(mol)
Theo PTHH, ta có nH2/nAl=3/2
-> nAl= 0,2(mol)
-> mAl= 0,2x27=5,4(g)
-> m=5,4(g)
c/ Theo pthh-> nH2/nHCl=3/6
-> nHCl= 0,6(mol)
-> mHCl= 0,6x(1+35,5)=21,9(g)
-> %HCl= 21,9/ 150x100%=14,6%
d/Theo pthh-> nH2/nAlCl3=3/2
-> nAlCl3= 0,2(mol)
-> mAlCl3= 0,2x(27+35.5x3)=26,7(g)
-> %AlCl3=26,7/(150-21,9+26,7)x100% ≈17,25%