8.00 mL of 5.00 x 103 M Fe(NO3)3 (5.00 times 10 to the minus 3rd power MF e (NO3)3) is added to 4.00 mL of 4.00 x 10-3 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

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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
Transcribed Image Text: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 (Fe3+) added into the solution and then we will calculate the number of moles of ferric ion present as [Fe(SCN)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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