Chemistry Forum

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!

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

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.

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