Question

Hòa tan 1 mol Kali vào 200g dd FeCl₃ 19.5% được dd A . Tính C%

Answer 1

mFeCl3= 200*19.5/100=39g

nFeCl3= 39/162.5=0.24 mol

K + H2O --> KOH + 1/2H2

1____________1______0.5

FeCl3 + 3KOH --> Fe(OH)3 + 3KCl

Bđ: 0.24_____1

Pư : 0.24_____0.72_____0.24_____0.72

Kt: 0________0.28______0.24____0.72

mFe(OH)3= 0.24*107=25.68g

mH2= 0.5*2=1g

mdd sau phản ứng = 39 + 200 - 1 - 25.68 = 212.32g

C%KOH dư = 15.68/212.32*100%= 7.38%

%KCl= 53.64/212.32*100%= 25.26%

Answer 2

2K + 2H₂O → 2KOH + H₂↑ (1)

3KOH + FeCl₃ → 3KCl + Fe(OH)₃↓ (2)

\(m_{FeCl_3}=200\times19,5\%=39\left(g\right)\)

\(\Rightarrow n_{FeCl_3}=\frac{39}{162,5}=0,24\left(mol\right)\)

Theo PT1: \(n_{KOH}=n_K=1\left(mol\right)\)

Theo PT2: \(n_K=3n_{FeCl_3}\)

Theo bài: \(n_K=\frac{25}{6}n_{FeCl_3}\)

Vì \(\frac{25}{6}>3\) ⇒ KOH dư

DD A gồm: KOH dư và KCl

\(m_K=1\times39=39\left(g\right)\)

Theo Pt1: \(n_{H_2}=\frac{1}{2}n_K=\frac{1}{2}\times1=0,5\left(mol\right)\)

\(\Rightarrow m_{H_2}=0,5\times2=1\left(g\right)\)

Theo PT2: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,24\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,24\times107=25,68\left(g\right)\)

Ta có: \(m_{dd}saupư=39+200-1-25,68=212,32\left(g\right)\)

Theo pT2: \(n_{KOH}pư=3n_{FeCl_3}=3\times0,24=0,72\left(mol\right)\)

\(\Rightarrow n_{KOH}dư=1-0,72=0,28\left(mol\right)\)

\(\Rightarrow m_{KOH}dư=0,28\times56=15,68\left(g\right)\)

\(\Rightarrow C\%_{KOH}dư=\frac{15,68}{212,32}\times100\%=7,39\%\)

Theo PT2: \(n_{KCl}=3n_{FeCl_3}=3\times0,24=0,72\left(mol\right)\)

\(\Rightarrow m_{KCl}=0,72\times74,5=53,64\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\frac{53,64}{212,32}\times100\%=25,26\%\)