gọi n Al= x mol
n Fe = y mol
n H2=$\frac{8,736}{22,4}$ =0,39 mol
2Al+6HCl→2AlCl3+3H2↑
x→ 1,5x mol
Fe+2HCl→FeCl2+H2↑
y → y mol
ta có hệ pt:27x+56y=11,58 x=0,18
1,5x+y=0,39 y=0,12
%m Al=$\frac{0,18.27.100}{11,58}$ ≈41,97 %
%m Fe=100%-41,97 %=58,03 %
2Al+ 6HCl→2AlCl3+3H2↑
0,18→0,54 mol
Fe+ 2HCl→FeCl2+H2↑
0,12 →0,24 mol
m ct HCl=(0,54+0,24).36,5=28,47 g
mdd HCl=$\frac{28,47.100}{36,5}$=78 g
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