\(n_{KOH}=\dfrac{200.8,4}{100}:56=0,3\left(mol\right)\)
\(n_{FeCl_3}=\dfrac{250.3,25}{100}:162,5=0,05\left(mol\right)\)
\(3KOH+FeCl_3\rightarrow Fe\left(OH\right)_3\downarrow+3KCl\)
0,15 <----- 0,05 -----> 0,05 --------> 0,15
Xét tỉ lệ thấy: \(\dfrac{0,3}{3}>\dfrac{0,05}{1}\) nên KOH dư sau phản ứng.
\(n_{KOH.dư}=0,3-0,15=0,15\left(mol\right)\)
Theo pthh \(n_{kt}=n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,05\left(mol\right)\)
\(\Rightarrow m=m_{Fe\left(OH\right)_3}=0,05.107=5,35\left(g\right)\)
Nung kết tủa:
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
0,05---------> 0,025
Theo pthh \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=\dfrac{1}{2}.0,05=0,025\left(mol\right)\)
\(q=m_{F_2O_3}=0,025.160=4\left(g\right)\)
Dung dịch X gồm \(\left\{{}\begin{matrix}KOH:0,15\left(mol\right)\\KCl:0,15\left(mol\right)\end{matrix}\right.\)
\(m_{dd.X}=m_{dd.KOH}+m_{dd.FeCl_3}-m_{Fe\left(OH\right)_3}=200+250-5,35=444,65\left(g\right)\)
\(C\%_{KOH}=\dfrac{0,15.56.100}{444,65}=1,89\%\)
\(C\%_{KCl}=\dfrac{0,15.74,5.100}{444,65}=2,51\%\)