a. PTHH: \(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow3Na_2SO_4+2Fe\left(OH\right)_3\\ \dfrac{1}{43}mol:\dfrac{6}{43}mol\rightarrow\dfrac{3}{43}mol:\dfrac{2}{43}mol\)
b. \(n_{NaOH}=\dfrac{16,8}{40}=0,42\left(mol\right)\)
\(n_{Fe\left(SO_4\right)_3}=\dfrac{8}{244}=\dfrac{1}{43}\left(mol\right)\)
Ta có tỉ lệ: \(\dfrac{0,42}{6}>\dfrac{1}{43}\)
Vậy \(Fe\left(SO_4\right)_3\) phản ứng hết.
\(m_{Fe\left(OH\right)_3}=\dfrac{2}{43}.107=4,98\left(g\right)\)
c. \(NaOH\) phản ứng còn dư.
\(m_{NaOHdu}=m_{NaOHbd}-m_{NaOHpu}\)
\(\Leftrightarrow m_{NaOHdu}=16,8-\left(\dfrac{6}{43}.40\right)=11,2\left(g\right)\)
d. PTHH: \(2Fe\left(OH\right)_3\underrightarrow{to}Fe_2O_3+3H_2O\\ \dfrac{2}{43}mol\rightarrow\dfrac{1}{43}mol:\dfrac{3}{43}mol\)
\(m_{Fe_2O_3}=\dfrac{1}{43}.160=3,72\left(g\right)\)
