a. \(n_{NaOH}=0,2mol;n_{FeCl_3}=0,2.0,6=0,12mol\)
\(3NaOH+FeCl_3->3NaCl+Fe\left(OH\right)_3\)
Lập tỷ lệ \(\dfrac{n_{NaOH}}{3}=\dfrac{0.2}{3}< \dfrac{n_{FeCl_3}}{1}=0,12\) => FeCl3 còn dư
\(n_{kếttủa}=n_{Fe\left(OH\right)_3}=\dfrac{1}{3}n_{NaOH}=\dfrac{0,2}{3}\left(mol\right)\)
\(m_{kt}=\dfrac{0.2}{3}.107=7,13g\)
b. \(V_{ddspu}=200+200=400\left(mL\right)\)