Sciencemadness Discussion Board

Neutralizing

mr40k - 11-7-2008 at 03:58

How much kmn04 to neutralize 1lb of h2s??? Can anyone answer this??? Cant seem to find the answer anywhere... I have 1000bbls of sour water that has 400PPM of H2S and I need to neutralize it... I figured out to be 139.9 lbs of H2S...

ScienceSquirrel - 11-7-2008 at 06:35

Could you clarify just how much water do you have?
1000bbls of water sounds like a lot...

mr40k - 11-7-2008 at 07:05

This is a sour water tank in a refinery, that I have been asked to neutralize. It has 1000 barrels in it. 42,000 gals. There are other alternatives but they want to use Potassium Permanganate(kMn04)..

ScienceSquirrel - 11-7-2008 at 07:33

Will you be adding sulphuric acid or does it have enough acid in it already?
Potassium permanagante is a better oxidiser in acid solution as it does not precipitate MnO2 due to incomplete oxidation.

ScienceSquirrel - 11-7-2008 at 07:39

Are these Imperial or US gallons?
You get a little bit more in an Imperial gallon and on this scale it will be loads!

mr40k - 11-7-2008 at 09:35

Its slightly acidic already and these are US gallons.. Ive done this type of job before but cant remember the quanity needed to neutralize the H2S... Already have figured there is 139.9 lbs of raw H2S..

12AX7 - 11-7-2008 at 09:46

K+ + MnO4- + 4 H2S + 3 H+ = K+ + Mn(2+) + 4 S + 4 H2O
There may be losses from S being oxidized to sulfates.

Atomic weights:
KMnO4: 39 + 55 + 4*16 = 158,
4 H2S: 4*(2*1 + 32) = 4*(34) = 136,
If the H+ comes from H2SO4, you will need 147.

So 140 lbs H2S is close enough to 136, so you'll need roughly 165 lbs KMnO4 and 150 lbs (about 12 gal.?) H2SO4 to complete this reaction.

The manganese sulfate solution remaining will be hazardous waste. Doing this reaction under neutral or basic conditions, yielding MnO2 primarily (and needing almost twice as much KMnO4) might be preferrable.

Tim

ScienceSquirrel - 11-7-2008 at 15:41

I am going to start by converting everything into SI units.

42,000 US gallons = n kg H20
1 US gallon = 3.79 kg H20

Hence n kg H2O = (42,000x3.79)/1 = 159,180 kg H2O
400 ppm H2S is equal to 400g of H2S per 1000kg of H2O

Thus;

400g H2S is contained in 1000kg H2O
xg H2S is contained in 159,180 kg H2O

x = (400x159,180)/1000

x= 63,672g

or in lbs

x = 63,672/(28.5x16)

x= 139.63lb which agrees with your result :P give or take a bit.

Time for a problem;

Take a careful look at 12AX7's equation, eleven hydrogen atoms on the left hand side and only eight on the right...

Time for a bit of homework :-)

Formatik - 11-7-2008 at 15:52

The equation:

2 KMnO4 + 5 H2S + 3 H2SO4 = 2 MnSO4 + K2SO4 + 5 S + 8 H2O from: Introduction to General Chemistry By Herbert Newby McCoy, Ethel Mary Terry (1920), pg. 381.

Would need:

Hint: this thread
but also:

Nitration calculations
Chlorate from pool supplies
etc.

And 1 pound = 453.59237 grams.

ScienceSquirrel - 11-7-2008 at 17:04

Quote:
Originally posted by Schockwave
The equation:

2 KMnO4 + 5 H2S + 3 H2SO4 = 2 MnSO4 + K2SO4 + 5 S + 8 H2O from: Introduction to General Chemistry By Herbert Newby McCoy, Ethel Mary Terry (1920), pg. 381.

Would need:

Hint: this thread
but also:

Nitration calculations
Chlorate from pool supplies
etc.

And 1 pound = 453.59237 grams.



I was going to tell them about that tomorrow!

:D