2Al + 6HCl-------> 2AlCl3 + 3H2(1)
Fe + 2HCl -------> FeCl2 + H2(2)
a. Ta có : n H2 ( đktc)=4,48/22,4=0,2 mol
Ta có Pt: 27x+56y=5,5
3/2x+y=0,2
=> x=0,1 mol
y=0,05 mol
=>m Al=27.0,1=2,7g
m Fe=0,05.56=2,8g
=>%m Al= 2,7/5,5.100%=49%
=>%m Fe=100%-49%=51%
b.Theo PTHH: n AlCl3=nAl=0,1mol
n FeCl2=n Fe=0,05 mol
(1) n HCl = 6/2 n Al=0,3mol
(2) n HCl=2nFe=0,1 mol
=>n HCl (2pt)=0,3+0,1=0,4 mol
=>m HCl=0,4.36,5=14,6g
=>mdd HCl=14,6.100%/14,6=100g
Ta có:
m AlCl3=0,1.133,5=13,35g
m FeCl2=0,05.127=6,35g
Theo Định luật btoan khlg:
m dd sau pứ=m hỗn hợp + mdd HCl-m H2
=5,5+100-(0,2.2)=105,1g
=> C% AlCl3=13,35/105,1.100%=12,7%
=>C% FeCl2=6,35/105,1.100%=6%