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July 4, 2006
Mobile Labs Could Not Have Produced Hydrogen As Described, Part I

In Part I of ChemicalConsultant's analysis of the mobile weapons laboratories, he calls into question the CIA's calculations of the production capability of the facilities described. In his calculations, he posits that these mobile facilities could not have produced the hydrogen necessary for the mission the CIA claims.

1. The reaction to produce hydrogen gas from aluminum, sodium hydroxide is:

2Al(s) +2NaOH (aq) +6H2O-> 2Na+ (aq) + 2[Al(OH)4]- +3H2 (g)

This means that it takes 80 grams of NaOH (molecular weight about 40) to make 6 grams of H2 (molecular weight about 2) and uses 54 grams of Al (atomic weight about 27) in the process. On a kilogram basis, 1 kg NaOH makes 6/80 = 0.075 kg or 75 g H2 and uses 54/80 = 0.675 kg or 675 g Al.

My reference is www.webelements.com/webelements/elements/text/Al/chem.html.

2. According to the Fast Facts link on the website of a major hydrogen producer, 1 kg of hydrogen gas at 1 atmosphere at 70o F occupies 11.986 cubic meters (m3) or 1 m3 weighs 1/ 11.896 = 0.084 kg or 84 g.

My reference is www.airproducts.com/products/fastfacts/charts_n_tables/32100/hydrogen.asp. Please note that the spaces on either side of n are underlined.

3. The Iraqi Survey Group Final Report, Annex D, Biological Weapons

Process Description section, states that “This is a batch process designed to produce sufficient H2 to fill 5x40l bottles to a pressure of between 40-50 bar”. This is equivalent to 50x5x40l =10,000 liters or 10 m3 at atmospheric pressure. Thus, from 2. , 0.84 kg of hydrogen must be produced. The next sentence in Process Description; the report states “This requires 10-12 kg of Aluminum powder, 1-1.5 kg flaked/ granulated NaOH, and 25-30 liters of water.” 1 kg of NaOH would only make 0.075 kg. However, from 1. , what is actually needed is 0.84/0.075 = 11.2 kg NaOH, using up 7.56 kg aluminum.

4. At first I thought that there might have been a typo. I went back to the Process Outline section of the report which described the purported Russian system which is one tenth the volume of the Iraqi reactor. I found that the ratio of reagents for the Russian system is the same as the report states for the Iraqi system. Thus the 100 g or 0.1 kg NaOH would only make 7.5 g hydrogen instead of the 84 g needed for the 1 m3 balloon. How is it that the “experts” who wrote the report and those who approved it did not catch these errors?

In part II tomorrow, ChemicalConsultant talks about how the residue in the mobile labs should not have been present and appear to be a deliberate ruse. He also shows how the formulas used for the Duelfer analysis would have produced hydrogen so impure as to be useless.

UPDATE: The molecular weight of NaOH is 40; thanks to those who pointed out the typo.

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Posted by Ed Morrissey at July 4, 2006 12:01 AM

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