nFe = \(\dfrac{11,2}{56}\)=0,2 ( mol )
mHCl = \(\dfrac{20.36,5}{100}\) = 7,3 (g)
=> nHCl = \(\dfrac{7,3}{36,5}\)= 0,2 ( mol )
Fe + 2HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)
0,1___0,2_____0,1___0,1 ( mol )
mdd sau phản ứng = 11,2 + 36,5 - 0,1.2 = 47,5
mFe dư = 56.(0,2-0,1) = 5,6 ( g )
C% Fe dư = \(\dfrac{5,6}{47,5}\).100 \(\approx\) 11,8%