Question

cho 200g dd Fe2(SO4)3 16% +300ml dd NaOH 2M (1,02g/ml)

sau pứ thu đc 1 kết tủa, lọc lấy kết tủa, nung nóng đến m ko đổi tạo thành a g chất rắn X

a PTHH

tính a g chất rắn X

b C% dd sau pứ

Answer 1

a) Fe₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Fe(OH)₃↓ (1)

2Fe(OH)₃ \(\underrightarrow{to}\) Fe₂O₃ + 3H₂O (2)

\(m_{Fe_2\left(SO_4\right)_3}=200\times16\%=32\left(g\right)\)

\(\Rightarrow n_{Fe_2\left(SO_4\right)_3}=\frac{32}{400}=0,08\left(mol\right)\)

\(n_{NaOH}=0,3\times2=0,6\left(mol\right)\)

Theo Pt1: \(n_{Fe_2\left(SO_4\right)_3}=\frac{1}{6}n_{NaOH}\)

Theo bài: \(n_{Fe_2\left(SO_4\right)_3}=\frac{2}{15}n_{NaOH}\)

Vì \(\frac{2}{15}< \frac{1}{6}\) ⇒ NaOH dư

Theo PT1: \(n_{Fe\left(OH\right)_3}=2n_{Fe_2\left(SO_4\right)_3}=2\times0,08=0,16\left(mol\right)\)

Theo Pt2: \(n_{Fe_2O_3}=\frac{1}{2}n_{Fe\left(OH\right)_3}=\frac{1}{2}\times0,16=0,08\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,08\times160=12,8\left(g\right)\)

Vậy \(a=12,8\left(g\right)\)

b) \(m_{ddNaOH}=300\times1,02=306\left(g\right)\)

\(m_{Fe\left(OH\right)_3}=0,16\times107=17,12\left(g\right)\)

Ta có: \(m_{dd}saupư=200+306-17,12=488,88\left(g\right)\)

Theo pT1: \(n_{NaOH}pư=6n_{Fe_2\left(SO_4\right)_3}=6\times0,08=0,48\left(mol\right)\)

\(\Rightarrow n_{NaOH}dư=0,6-0,48=0,12\left(mol\right)\)

\(\Rightarrow m_{NaOH}dư=0,12\times40=4,8\left(g\right)\)

\(\Rightarrow C\%_{NaOH}dư=\frac{4,8}{488,88}\times100\%=0,98\%\)

Theo PT1: \(n_{Na_2SO_4}=3n_{Fe_2\left(SO_4\right)_3}=3\times0,08=0,24\left(mol\right)\)

\(\Rightarrow m_{Na_2SO_4}=0,24\times142=34,08\left(g\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\frac{34,08}{488,88}\times100\%=6,97\%\)