  Suppose a student mixes 5.00 mL of 2.00 x 10-3 M Fe(NO,)3 with 4.00 mL of 2.00 x 10-3 MKSCN and 1.00 mL of water. The student then determines the [FeNCS2+] at equilibrium to be9.25 X 10 M. Find the equilibrium constant for the following reaction. Show all your calcula-tions for each step.3/3Fe3+(aq)SCN (aq FeNCS2+(aq)Step 1. Calculate the initial number of moles of Fe3+ and SCN (use Equation 12).moles of Fe3+moles of SCNStep 2. How many moles of FeNCS2+ are present atequilibrium? What is the volume of theequilibrium mixture?mLmoles of FeNCS2+How many moles of Fe3 and SCN are consumed to produce the FeNCS2+?moles of Fe3+moles of SCNStep 3. Calculate the number of moles of Fe3+ and SCN- remaining at equilibrium. Use Equa-tion (13) and the results of Steps 1 and 2.moles of Fe3+moles of SCN9

Question

I have a 3 step problem about Determination of an Equilbrium constant. help_outlineImage TranscriptioncloseSuppose a student mixes 5.00 mL of 2.00 x 10-3 M Fe(NO,)3 with 4.00 mL of 2.00 x 10-3 M KSCN and 1.00 mL of water. The student then determines the [FeNCS2+] at equilibrium to be 9.25 X 10 M. Find the equilibrium constant for the following reaction. Show all your calcula- tions for each step. 3/3 Fe3+ (aq) SCN (aq FeNCS2+ (aq) Step 1. Calculate the initial number of moles of Fe3+ and SCN (use Equation 12). moles of Fe3+ moles of SCN Step 2. How many moles of FeNCS2+ are present at equilibrium? What is the volume of the equilibrium mixture? mL moles of FeNCS2+ How many moles of Fe3 and SCN are consumed to produce the FeNCS2+? moles of Fe3+ moles of SCN Step 3. Calculate the number of moles of Fe3+ and SCN- remaining at equilibrium. Use Equa- tion (13) and the results of Steps 1 and 2. moles of Fe3+ moles of SCN 9 fullscreen
Step 1

Step – 1: The initial moles of given Fe3+ and SCN- ions are determined as follows, help_outlineImage Transcriptionclosemoles of solute Molarity= Volume of solution in L moles of Fe3 Molarity x Volume of solution in L 11 -2 x 103 M x 5 x103 L = 1x105 moles moles of SCN = Molarity x Volume of solution in L =2 x 103 M x 4 x103 L = 8 x106 moles fullscreen
Step 2

Step – 2: The moles of FeNCS2+ at equilibrium and the equili... help_outlineImage TranscriptioncloseVolume of equilibrium mixture 5 mL+4mL =9 mL moles of solute Molarity= Volume of solution in L moles of FeNCS2 Molarity x Volume of solution in L 9.25 x 105 M x 9 x 103 L = 8.325x 107 moles Consumed reactants moles moles of FeNCs produced at equilibrium Fe and SCN moles consumed 8.325x 10 moles fullscreen

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