Project:Deactivating a ferric chloride solution: Difference between revisions

Project:Deactivating a ferric chloride solution (view source)

Revision as of 21:05, 28 May 2010

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→‎In Theory

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 Chemically the reaction is Ferric Chloride reacting to Sodium Hydroxide turning into Ferric Hydroxide and Sodium Chloride.
-FeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl
+FeCl<sub>3</sub> + 3 NaOH → Fe(OH)<sub>3</sub> + 3 NaCl
 The proportions are:
-mole of FeCl3 which is 91,3g reacting with 3 moles of Sodium Hydroxide 3NaOH which is 120g.
+mole of FeCl<sub>3</sub> which is 91,3g reacting with 3 moles of Sodium Hydroxide 3 NaOH which is 120g.
 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).
-Ferric Chloride (FeCl3) will then have the molecular mass of 55.8 * 1 + 35.5 * 3 = 162.3 g/mole.
+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 Ferric Chloride the calculation is 300/162.3 = 1.85 moles of Ferric Chloride.
-Each mole of FeCl3 requires 3 moles of NaOH (Sodium Hydroxide) at 40 grams per mole.
+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 FeCl3 solution is obtained thus:
+The weight of Na0H required to deactivate the spent FeCl<sub>3</sub> solution is obtained thus:
-.85 * 3 = 5.55 moles of NaOH, at 40 g/mole that will be 5.55 * 40 = 222 grams NaOH
+.85 * 3 = 5.55 moles of NaOH, at 40 g/mole that will be 5.55 * 40 = 222 grams NaOH.