9.00 mL of 9.00 x 10-³ M Fe(NO3)3 (9.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 4.00 mL of 7.00 x 10-3 M KSCN (7.00 times 10 to the minus 3rd power M KSCN) along with 4.00 mL of water. The concentration of [F e(SCN)²+] was found to be 1.00 x 10-4 M (1.00 times 10 to the minus 4th power M) at equilibrium. How many moles of Fe³+ are present in the solution at equilibrium?

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9.00 mL of 9.00 x 10-3 M Fe(NO3)3 (9.00 times 10 to the minus 3rd power M F e
(NO3)3) is added to 4.00 mL of 7.00 x 10-3 M KSCN (7.00 times 10 to the minus 3rd
power M KSCN) along with 4.00 mL of water. The concentration of [F e(SCN)²+] was
found to be 1.00 x 10-4 M (1.00 times 10 to the minus 4th power M) at
equilibrium.How many moles of Fe³+ are present in the solution at equilibrium?
Transcribed Image Text:9.00 mL of 9.00 x 10-3 M Fe(NO3)3 (9.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 4.00 mL of 7.00 x 10-3 M KSCN (7.00 times 10 to the minus 3rd power M KSCN) along with 4.00 mL of water. The concentration of [F e(SCN)²+] was found to be 1.00 x 10-4 M (1.00 times 10 to the minus 4th power M) at equilibrium.How many moles of Fe³+ are present in the solution at equilibrium?
Expert Solution
Step 1: Basic discussion

We will first calculate the millimoles of all the reactants added and the millimoles of the final product formed. After that, we will use the balance reaction to find out the millimoles of Fe3+ used in the process as the equilibrium concentration of Fe(SCN)2+ is already given.

Equilibrium mmol of Fe3+ is-

equals Initial space mmol space of space Fe to the power of 3 plus end exponent space minus space mmol space of space Fe to the power of 3 plus end exponent space reacted


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