Project:Deactivating a ferric chloride solution: Difference between revisions

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(Created page with 'Project maintained by Daniel Sikar ~/ dsikaratgmaildotcom. == Deactivating a Ferric Chloride solution == === In practice === * Example A solution that was made with a packet o…')
 
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To find how many moles are contained in 300 grams of Ferric Chloride the calculation is 300/162.3 = 1.85 moles of Ferric Chloride.
To find how many moles are contained in 300 grams of Ferric Chloride the calculation is 300/162.3 = 1.85 moles of Ferric Chloride.


These will require 1.85 * 3 = 5.55 moles of NaOH, at 40 g/mole that will be 5.55 * 40 = 222 grams NaOH
Each mole of FeCl3 requires 3 moles of NaOH (Sodium Hydroxide) at 40 grams per mole.


contains 1.85 moles. Each mole requires 3 moles of NaOH (Sodium Hydroxide) at 40 grams per mole.
The weight of Na0H required to deactivate the spent FeCl3 solution is obtained thus:
 
1.85 * 3 = 5.55 moles of NaOH, at 40 g/mole that will be 5.55 * 40 = 222 grams NaOH
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