\(n_{FeCl_2}=\dfrac{31,75}{127}=0,25\left(mol\right)\)
\(m_{KOH}=\dfrac{200\cdot8,4}{100}=16,8g\)
\(n_{KOH}=\dfrac{56}{16,8}=0,3\left(mol\right)\)
\(FeCl_2+2KOH\rightarrow Fe\left(OH\right)_2\downarrow+2KCl\)
1-------------2-------------1
0,25---------0,3
0,15---------0,3------------0,15
0,05----------0-------------0,15
\(m_{Fe\left(OH\right)_2}=0,05\cdot90=4.5g\)
\(m_{dd_{ }sau_{ }pu_{ }}=200+31,75-4,5=227,25\left(g\right)\)
\(C\%_{FeCl_2du}=\dfrac{0,05\cdot127}{227,25}\cdot100=2,8\%\)