a, PTHH: Fe+2HCl--->FeCl2+H2
nH2= \(\dfrac{0,224}{22,4}=0,01\) mol
Theo pt: nFe=nH2=0,01 mol
=> mFe= 0,01.56= 0,56 g
mCu= 1,2-0,56= 0,64 g
=> %Fe= \(\dfrac{0,56}{1,2}.100\%\approx46,7\%\)
%Cu= 100%-46,7%= 53,3%
b, Theo pt: nHCl=2.nH2= 2.0,01=0,02 mol
=> mHCl= 0,02.36,5= 0,73 g
=> C%ddHCl= \(\dfrac{0,73}{10}.100\%=7,3\%\)