\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=0,2.1,5=0,3\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
LTL: \(0,1< \dfrac{0,3}{2}\) => HCl dư
Theo pthh: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Fe}=2.0,1=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(HCl.dư\right)}=\dfrac{0,3-0,2}{0,2}=0,5M\end{matrix}\right.\)
\(b,m_{H_2O\left(hiđrat\right)}=\dfrac{14,3}{286}.10.18=9\left(g\right)\\ m_{H_2O\left(dd\right)}=9+35,7=45,7\left(g\right)\\ V_{dd}=\dfrac{45,7}{1}=45,7\left(l\right)\\ m_{Na_2CO_3}=14,3-9=5,3\left(g\right)\\ m_{dd}=14,3+35,7=50\left(g\right)\\ n_{Na_2CO_3}=\dfrac{5,3}{106}=0,05\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{5,3}{50}.100\%=10,6\%\\C_{M\left(Na_2CO_3\right)}=\dfrac{0,05}{45,7}=0,001M\end{matrix}\right.\)
