a) PTHH: 3KOH + FeCl3 \(\rightarrow\) 3KCl + Fe(OH)3\(\downarrow\)(1)
2Fe(OH)3 \(\underrightarrow{t^o}\) Fe2O3 + 3H2O (2)
b) mKOH = \(\frac{300.5,6}{1000}\) = 1,68(g)
=> nKOH = \(\frac{1,68}{56}=0,03\left(mol\right)\)
Theo PT (1) : n\(Fe\left(OH\right)_3\) = \(\frac{1}{3}n_{KOH}\) = \(\frac{1}{3}.0,03=0,01\left(mol\right)\)
Theo PT(2): n\(Fe_2O_3\) = \(\frac{1}{2}n_{Fe\left(OH\right)_3}\) = \(\frac{1}{2}.0,01=0,005\left(mol\right)\)
=> m\(Fe_2O_3\) = 0,005.160 = 0,8 (g) = mB
\(\text{a, 3KOH + F e C l 3 → 3KCl + F e ( O H ) 3 (1)}\)
\(\text{ 2 F e ( O H ) 3 → F e 2 O 3 +3 H 2 O (2)}\)
Ta có :
\(m_{KOH}=\frac{5,6.300}{100}\text{= 16.8 g}\)
\(\Rightarrow n_{KOH}\approx\text{ 0,285 mol}\)
Theo pthh (1) : \(n_{Fe\left(OH\right)3}=\frac{1}{3}n_{KOH}\text{= 0,095 mol}\)
Theo pthh (2) : \(n_{Fe2O3}=\frac{1}{2}n_{Fe\left(OH\right)3}=\text{0,0475 mol}\)
\(\text{⇒ m F e 2 O 3 = 7,6g}\)
