Project:Deactivating a ferric chloride solution: Difference between revisions
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Project:Deactivating a ferric chloride solution (view source)
Revision as of 10:59, 29 May 2010
, 29 May 2010→In Theory
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=== In Theory === | === In Theory === | ||
Ferric chloride reacts with sodium hydroxide turning into ferric hydroxide and sodium chloride. | |||
FeCl<sub>3</sub> + 3 NaOH → Fe(OH)<sub>3</sub> + 3 NaCl | FeCl<sub>3</sub> + 3 NaOH → Fe(OH)<sub>3</sub> + 3 NaCl | ||
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The proportions are: | The proportions are: | ||
1 mole of FeCl<sub>3</sub> which is 91,3g reacting with 3 moles of | 1 mole of FeCl<sub>3</sub> which is 91,3g reacting with 3 moles of NaOH which is 120g. | ||
The mass of 1 mole is obtained by adding together the atomic masses obtained from the periodic table of elements: | The mass of 1 mole is obtained by adding together the atomic masses obtained from the periodic table of elements: | ||
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Chlorine (Cl) has the atomic mass of 35,5 (grams/mole) and Iron (Fe) has the atomic mass of 55.8 (grams/mole). | Chlorine (Cl) has the atomic mass of 35,5 (grams/mole) and Iron (Fe) has the atomic mass of 55.8 (grams/mole). | ||
Ferric | Ferric chloride (FeCl<sub>3</sub>) will then have the molecular mass of 55.8 * 1 + 35.5 * 3 = 162.3 g/mole. | ||
To find how many moles are contained in 300 grams of | 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. | ||
Each mole of FeCl<sub>3</sub> requires 3 moles of NaOH ( | Each mole of FeCl<sub>3</sub> requires 3 moles of NaOH (sodium hydroxide) at 40 grams per mole. | ||
The weight of Na0H required to deactivate the spent FeCl<sub>3</sub> solution is obtained thus: | The weight of Na0H required to deactivate the spent FeCl<sub>3</sub> 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. | 1.85 * 3 = 5.55 moles of NaOH, at 40 g/mole that will be 5.55 * 40 = 222 grams NaOH. | ||
