Hello, I have two questions:
1. We want to prepare a 20.0% KOH solution. We have at our disposal a 9.96% KOH solution with a density of 1.090 g mL–1
(T = 20 °C) and a 32.0% KOH solution with a density of 1.310 g mL–1(T = 20 °C).
a.) Calculate in what volume ratio we must mix the 9.96% and 32.0% solution KOH to make a 20.0% KOH solution.
b) Calculate the molality of the solution.
2. Balance the chemical reaction equation: H2S + KMnO4 = K2SO4 + K2S2O3 + MnS + H2O + S 📖
Un grand merci d'avance!
Questions from general chemistry - mixing of solutions and balancing of equations of redox reactions
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- ChenBeier
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Re: Questions from general chemistry - mixing of solutions and balancing of equations of redox reactions
Cross rule
32.................10,04 weight part
..........20
9,96................12 weight part
_________________
.......................22,04
10,04 is equal 7,66 l 32%
and 12 is equal 11 l 9,96%
In sum we get 18,66 l of 20%.
The redoxreaction make no sense, because K2S2O3 also will oxidised to sulfate
Redoxpair H2S/ H2SO4
And MnO4- / Mn 2+
Oxidation H2S + 4 H2O => H2SO4 + 8 H+ + 8 e-
Reduction MnO4 - + 8 H+ + 5 e- => Mn 2+ + 4 H2O
KgV = 40 e-
5 H2S + 20 H2O => 5 H2SO4 + 40 H+ + 40 e-
8 MnO4 - + 64 H+ + 40 e- => 8 Mn 2+ + 32 H2O
Addition 5 H2S + 8 MnO4- + 24 H+ => 8 Mn 2+ + 5 H2SO4 + 12 H2O
Add K+
5 H2S + 8 KMnO4 + 24 H+ => 8 Mn 2+ + 8K+ 5 H2SO4 + 12 H2O
Add more H2S
17 H2S + 8 KMnO4 => 8 MnS + 4 K2S + 5 H2SO4 + 12 H2O
Reaction of sulfuric acid
13 H2S + 8 KMnO4 => 8 MnS + 4 K2SO4 + H2SO4 + 12 H2O
16 H2S + 8 KMnO4 => 8 MnS + 4 K2SO4 + 16 H2O + 4 S
Final
4 H2S + 2 KMnO4 => 2 MnS + K2SO4 + 4 H2O + S
32.................10,04 weight part
..........20
9,96................12 weight part
_________________
.......................22,04
10,04 is equal 7,66 l 32%
and 12 is equal 11 l 9,96%
In sum we get 18,66 l of 20%.
The redoxreaction make no sense, because K2S2O3 also will oxidised to sulfate
Redoxpair H2S/ H2SO4
And MnO4- / Mn 2+
Oxidation H2S + 4 H2O => H2SO4 + 8 H+ + 8 e-
Reduction MnO4 - + 8 H+ + 5 e- => Mn 2+ + 4 H2O
KgV = 40 e-
5 H2S + 20 H2O => 5 H2SO4 + 40 H+ + 40 e-
8 MnO4 - + 64 H+ + 40 e- => 8 Mn 2+ + 32 H2O
Addition 5 H2S + 8 MnO4- + 24 H+ => 8 Mn 2+ + 5 H2SO4 + 12 H2O
Add K+
5 H2S + 8 KMnO4 + 24 H+ => 8 Mn 2+ + 8K+ 5 H2SO4 + 12 H2O
Add more H2S
17 H2S + 8 KMnO4 => 8 MnS + 4 K2S + 5 H2SO4 + 12 H2O
Reaction of sulfuric acid
13 H2S + 8 KMnO4 => 8 MnS + 4 K2SO4 + H2SO4 + 12 H2O
16 H2S + 8 KMnO4 => 8 MnS + 4 K2SO4 + 16 H2O + 4 S
Final
4 H2S + 2 KMnO4 => 2 MnS + K2SO4 + 4 H2O + S
Re: Questions from general chemistry - mixing of solutions and balancing of equations of redox reactions
I appreciate your work, but I'm afraid I don't entirely comprehend how you arrived at the conclusion. What is the origin of the numerals 10,4 and 12? I'm also curious in the redox process and how it would balance, say, if it had hypothetical significance. In the book where the reaction was written, the solution is also given, but I have no idea how to get there or why a reaction balancing computer program can't balance it.
- ChenBeier
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- Joined: Wed Sep 27, 2017 7:25 am
- Location: Berlin, Germany
Re: Questions from general chemistry - mixing of solutions and balancing of equations of redox reactions
Cross rule check here
https://bodheeprep.com/mixture-and-alligation
Given solutions 32% and 9,96 % , target is 20%
Above write the larger one here 32% below the lower one 9,96%
In the middle of the cross the target 20%.
Build difference to get weight part
32 - 20 = 12 and 20 - 9,96 = 10,04
Convert to volume by using specific gravity.
The redox reaction can also developed to K2S2O3 , but in practise this compound is unstable.
Forget a Computer program
2 H2S + 3 H2O => S2O3 2- + 10 H+ + 8 e-
Try yourself the next steps
If you have the results add it to the other equation
https://bodheeprep.com/mixture-and-alligation
Given solutions 32% and 9,96 % , target is 20%
Above write the larger one here 32% below the lower one 9,96%
In the middle of the cross the target 20%.
Build difference to get weight part
32 - 20 = 12 and 20 - 9,96 = 10,04
Convert to volume by using specific gravity.
The redox reaction can also developed to K2S2O3 , but in practise this compound is unstable.
Forget a Computer program
2 H2S + 3 H2O => S2O3 2- + 10 H+ + 8 e-
Try yourself the next steps
If you have the results add it to the other equation