Đặt:
nMg= x mol
nAl= y mol
mhh= 24x + 27y = 7.8g (1)
Mg + 2HCl --> MgCl2 + H2
x_____________x______x
2Al + 6HCl --> 2AlCl3 + 3H2
y______________y______1.5y
nH2= x + 1.5y = 0.8/2=0.4 mol (2)
Giải (1) và (2) :
x= 0.1
y= 0.2
mM= mMgCl2 + mAlCl3= 0.1*95 + 0.2*133.5= 36.2g
b)
MgCl2 + 2KOH --> Mg(OH)2 + 2KCl
0.1_______0.2
AlCl3 + 3KOH --> Al(OH)3 + 3KCl
0.2______0.6
VddKOH= (0.2+0.6)/2=0.4l