\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{HCl}=0.12\cdot2=0.24\left(mol\right)\)
\(Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\uparrow\)
0.1 → 0.2 → 0.1
\(\Rightarrow n_{HCl\left(\text{dư}\right)}=0.24-0.2=0.04\left(mol\right)\)
\(C_{M_{HCl\left(\text{dư}\right)}}=\dfrac{0.04}{0.12}=\dfrac{1}{3}\left(M\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0.1}{0.12}=\dfrac{5}{6}\left(M\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,12.2=0,24\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
trc p/u : 0,1 0,24
p/u : 0,1 0,2 0,1 0,1
sau : 0 0,04 0,1 0,1
-> HCl dư sau p/ư
\(C_{M_{FeCl_2}}=\dfrac{0,1}{0,12}=\dfrac{5}{6}\left(M\right)\)
\(C_{M_{HCldư}}=\dfrac{0,04}{0,12}=\dfrac{1}{3}\left(M\right)\)