\(n_{BaOH}=\dfrac{300.30\%}{154}=0,58\left(mol\right)\)
Pt: \(2BaOH+FeSO_4\rightarrow Ba_2SO_4+Fe\left(OH\right)_2\)
0,58mol \(\rightarrow0,29mol\) \(\rightarrow0,29mol\)
\(m_{Ba_2SO_4}=0,29.370=107,3\left(g\right)\)
\(\Sigma_{m_{dd}\left(spu\right)}=300+800-107,3=992,7\left(g\right)\)
\(C\%_{Fe\left(OH\right)_2}=\dfrac{0,29.90}{992,7}.100=2,63\%\)