Đáp án:
\(\begin{array}{l} b,\ C_{M_{HCl}}=1,5\ M.\\ c,\ m_{KOH}\text{(dư)}=5,6\ g.\\ m_{KCl}=22,35\ g.\\ m_{Fe(OH)_2}=13,5\ g.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:Fe+2HCl\to FeCl_2+H_2↑\\ b,\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\ mol.\\ Theo\ pt:\ n_{HCl}=2n_{H_2}=0,3\ mol.\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\ M.\\ Theo\ pt:\ n_{FeCl_2}=n_{H_2}=0,15\ mol.\\ c,\ PTHH:FeCl_2+2KOH\to Fe(OH)_2↓+2KCl\\ n_{KOH}=0,2\times 2=0,4\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,15}{1}<\dfrac{0,4}{2}\\ \Rightarrow KOH\ \text{dư.}\\ \Rightarrow n_{KOH}\text{(dư)}=0,4-(0,15\times 2)=0,1\ mol.\\ Theo\ pt:\ n_{KCl}=2n_{FeCl_2}=0,3\ mol.\\ Theo\ pt:\ n_{Fe(OH)_2}=n_{FeCl_2}=0,15\ mol.\\ \Rightarrow m_{KOH}\text{(dư)}=0,1\times 56=5,6\ g.\\ m_{KCl}=0,3\times 74,5=22,35\ g.\\ m_{Fe(OH)_2}=0,15\times 90=13,5\ g.\end{array}\)