\(n_{Fe\left(NO_3\right)_3}=\dfrac{48,4}{242}=0,2mol\)
\(n_{NaOH}\dfrac{32}{40}=0,8mol\)
Fe(NO3)3+3NaOH\(\rightarrow\)Fe(OH)3\(\downarrow\)+NaNO3
- Tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\)
nên NaOH dư
\(n_{Fe\left(OH\right)_3}=n_{Fe\left(NO_3\right)_3}=0,2mol\)
mY=0,2(56+17.3)=21,4 gam
mddY=48,4+32+400-21,4=458,8 gam