1.) Kc for the reaction of tin (ii) ions with iron (iii) ions Sn2+ (aq) +2FE3+ (aq) Snt+ (aq) +2FE2* (aq) is 1.0 x1010 at 298K. In an experiment at room temperature in which solutions of Sn2+ (aq) and Fe3+ (aq) were mixed, the equilibrium concentrations of tin ions were found to be [Sn2+ (aq)] = 0.050 mol dm-3 Sn4+ (aq)] = 0.040 mol dm-3 Calculate the concentrations of iron(iii) and iron(ii) ions at equilibrium.
1.) Kc for the reaction of tin (ii) ions with iron (iii) ions Sn2+ (aq) +2FE3+ (aq) Snt+ (aq) +2FE2* (aq) is 1.0 x1010 at 298K. In an experiment at room temperature in which solutions of Sn2+ (aq) and Fe3+ (aq) were mixed, the equilibrium concentrations of tin ions were found to be [Sn2+ (aq)] = 0.050 mol dm-3 Sn4+ (aq)] = 0.040 mol dm-3 Calculate the concentrations of iron(iii) and iron(ii) ions at equilibrium.
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section12.5: Using Equilibrium Constants
Problem 12.6PSP
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![1.) Kc for the reaction of tin (i) ions with iron (iii) ions
Sn2+ (ag) +2FE3+ (aq) = Sn+ (aq) +2FE2+ (aq) is 1.0 x1010 at
298K. In an experiment at room temperature in which solutions of
Sn2+ (aq) and Fe3+ (ag) were mixed, the equilibrium concentrations of
tin ions were found to be
[Sn²+ (aq)] = 0.050 mol dm³
Sn4+ (aq)] = 0.040 mol dm-3
Calculate the concentrations of iron(iii) and iron(ii) ions at equilibrium.
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Transcribed Image Text:1.) Kc for the reaction of tin (i) ions with iron (iii) ions
Sn2+ (ag) +2FE3+ (aq) = Sn+ (aq) +2FE2+ (aq) is 1.0 x1010 at
298K. In an experiment at room temperature in which solutions of
Sn2+ (aq) and Fe3+ (ag) were mixed, the equilibrium concentrations of
tin ions were found to be
[Sn²+ (aq)] = 0.050 mol dm³
Sn4+ (aq)] = 0.040 mol dm-3
Calculate the concentrations of iron(iii) and iron(ii) ions at equilibrium.
%3D
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