Hydrogen Fluoride gas mixture quantities

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SUMMARY

The discussion centers on calculating the quantity of hydrogen fluoride (HF) produced from a mixture of 17 tons of hydrogen and 8 tons of fluorine, relevant to the European Seveso directives for lower tier sites. The calculation indicates that fluorine is the limiting reactant, resulting in the formation of 8.4 tons of HF. This quantity is crucial for determining compliance with Seveso regulations, as it could potentially push the total above the upper threshold of 20 tons. The formula used for the calculation is based on the atomic mass of fluorine and its reaction with hydrogen.

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  • Understanding of chemical reactions and stoichiometry
  • Familiarity with the European Seveso directives
  • Basic knowledge of atomic mass and molecular weight calculations
  • Awareness of safety regulations concerning hazardous materials
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Rudi Rose
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Hi,

I am not a chemist but maybe there is somebody who can help me.
I need a put together an emergency plan as part of a college law assignment on European Seveso directives.
My case study has 17t of Hydrogen and 8t of Fluorine.
Both of which combine satisfy the requirement for a lower tier Seveso site.
8 (Quantity)/10 (threshold limit of fluorine) + 17/5 (threshold limit of hydrogen)...greater than 1 @4.2 makes it a lower tier site.
However if combined they create hydrogen fluoride which may push them over the upper threshold of 20t.
So my question how much hyrdrogen fluorine would be created with 17t of hydrogen ane 8t of fluorine.

Thanks

R
 
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Fluorine would be the limiting factor, with 19u/atom it catches a maximum of 1/19 of its mass as hydrogen to form 20/19*8t = 8.4 tons of HF.
 
Thanks mfb