Ta co: \(\dfrac{0,2}{3}< \dfrac{0,3}{1}\Rightarrow\) FeCl3 dư
3KOH + FeCl3 \(\rightarrow\) 3KCl + Fe(OH)3
de: 0,2 0,3
pu: 0,2 \(\dfrac{1}{15}\) 0,2 \(\dfrac{1}{15}\)
spu: 0 \(\dfrac{7}{30}\) 0,2 \(\dfrac{1}{15}\)
\(m_{FeCl_3\left(dư\right)}=\dfrac{7}{30}.162,5\approx37,92g\)
\(m_{KCl}=0,2.74,5=14,9g\)
\(m_{Fe\left(OH\right)_3}=\dfrac{1}{15}.107\approx7,13g\)