8.00 mL of 5.00 x 10³ MFe(NO3)3 (5.00 times 10 to the minus3rd power MFe (NO3)3) is added to4.00 mL of 4.00 x 103 M KSCN(4.00 times 10 to the minus 3rd powerM KSCN) along with 6.00 mL of water.The concentration of [F e(SCN)2+] wasfound to be 2.00 x 10-4 M (2.00 times10 to the minus 4th power M) atequilibrium.How many moles of F e3+are present in the solution atequilibrium? Express your answer as adecimal number (no exponents).Your Answer:Answerunits

Answered: 8.00 mL of 5.00 x 103 M Fe(NO3)3 (5.00…

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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8.00 mL of 5.00 x 10³ M
Fe(NO3)3 (5.00 times 10 to the minus
3rd power MFe (NO3)3) is added to
4.00 mL of 4.00 x 103 M KSCN
(4.00 times 10 to the minus 3rd power
M KSCN) along with 6.00 mL of water.
The concentration of [F e(SCN)2+] was
found to be 2.00 x 10-4 M (2.00 times
10 to the minus 4th power M) at
equilibrium.How many moles of F e3+
are present in the solution at
equilibrium? Express your answer as a
decimal number (no exponents).
Your Answer:
Answer
units

Expert Solution

Step 1

Answer:

First of all we will calculate the number of moles of ferric ion $(F e^{3 +})$ added into the solution and then we will calculate the number of moles of ferric ion present as $[F e {(S C N)}^{2 +}]$ ions. Their difference will give us the value of number of moles of ferric ions present in the solution at equilibrium.

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