a)
\(n_{HCl}=\dfrac{43,8.20\%}{36,5}=0,24\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,12<-0,24---->0,12-->0,12
\(\Rightarrow\left\{{}\begin{matrix}V=0,12.22,4=2,688\left(l\right)\\a=0,12.56=6,72\left(g\right)\end{matrix}\right.\)
b) mdd sau pư = 6,72 + 43,8 - 0,12.2 = 50,28 (g)
\(C\%_{FeCl_2}=\dfrac{0,12.127}{50,28}.100\%=30,31\%\)