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The reaction is as follow: 4 KOH(aq) + FeCl4(aq) 4KCI(aq) + Fe(OH)4(s) Calculate the molarity of the non-spectator ion that didn't run out (that's not zero). Said another way, of the two non-spectator ions, calculate the left-over concentration of the non-limiting reactant.

The reaction is as follow: 4 KOH(aq) + FeCl4(aq) 4KCI(aq) + Fe(OH)4(s) Calculate the molarity of the non-spectator ion that didn't run out (that's not zero). Said another way, of the two non-spectator ions, calculate the left-over concentration of the non-limiting reactant.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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The following two solutions are mixed: 348.0 mL of 0.465 M KOH(aq) is combined with 18.64 mL of 1.827 M FeCla(aq). The reaction is as follow: 4 KOH(aq) + FeClą(aq) 4KCI(aq) + Fe(OH)4(s) Calculate the molarity of the non-spectator ion that didn't run out (that's not zero). Said another way, of the two non-spectator ions, calculate the left-over concentration of the non-limiting reactant.
Transcribed Image Text:The following two solutions are mixed: 348.0 mL of 0.465 M KOH(aq) is combined with 18.64 mL of 1.827 M FeCla(aq). The reaction is as follow: 4 KOH(aq) + FeClą(aq) 4KCI(aq) + Fe(OH)4(s) Calculate the molarity of the non-spectator ion that didn't run out (that's not zero). Said another way, of the two non-spectator ions, calculate the left-over concentration of the non-limiting reactant.
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