n(NaOH)=0,2 mol
PTHH: HCl+NaOH---> NaCl+H2O
=> n(HCl)=0,2 mol.
n(HCl trong dd)=1 mol=> n(HCl pứ)=0,8 mol
nAl=x; nFe=y
2Al+6HCl--->2AlCl3+3H2
x........3x........................1,5x
Fe+2HCl---> FeCl2+H2
y........2y.....................y
Ta có: 27x+56y=11; 3x+2y=0,8
=> x=0,2;y=0,1
=> V=8,96(l)
=>mAl=5,4; mFe=5,6
#Walker