a,
\(Fe+2HCl\rightarrow FeCl2+H_2\)
b + c ,
\(n_{Fe}=0,015\left(mol\right)=n_{FeCl2}\)
\(n_{HCl}=0,03\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}CM_{HCl}=0,3M\\CM_{FeCl2}=0,15M\end{matrix}\right.\)
n Fe= 0.84/56=0.015 (mol)
a) PTHH:
_Fe + 2HCl -> FeCl2 + H2
0.015->0.03__0.015_______(mol)
b) \(C_{M_{HCl\text{ đã dùng}}}\)=0.3/0.1=3M
c) \(C_{M_{FeCl_2}}\)=0.015/0.1=0.15M