4.
Fe+2hcl->fecl2+h2
x.....2x.....x..........x
2Al+6hcl>2alcl3+3h2
y.......3y.....y...........1,5y
Co
56x+27y=5,5
x+1,5y=0,2
=>x=0,05;y=0,1
mFe=56.0,05=2,8g
%mFe=2,8/5,5.100%=50,9%
%mAl=100%-50,9%=49,1%
mmuoi=0,05.127+0,1.133,5=19,7g
nHCl pứ=0,05.2+3.01=0,4mol
nHCl bđ=110/100.0,4=0,44mol
Vdd...