# chem please help! urgent?

Calculate the equilibrium concentration of [FeSCN2+] if 2.0mL of 2.0x10-3M KSCN and 25.00mL of 0.0020M Fe(NO3)3 are diluted to 100.0mL.

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- Bobby_ThinLv 71 month ago
calculate the moles that we start with:

(25.00 mL)*(2.00 x 10^-3 M Fe(NO3)3) = 0.0500 mmoles Fe(NO3)3 in 100mL

or 0.500 mmole/ liter

(2.0 mL)*(2.00 x 10^-3 M KSCN) = 0.004 moles of KSCN = .04 mmole/ liter

Fe3+ + SCN^-1---> Fe(SCN)^+2

from the equation we see .04 mmole of SCN- reacts with .04 mmole of Fe3+

to produce .04 mmol of Fe(SCN)+2 with SCN- as the limiting reactant

the Keq of Fe(SCN)^+2 is very large and the equilibrium lies almost completely to the RHS

[Fe(SCN)^2+] = .04 mmol / liter or 0.04 *10^-3 M or

4 .0 * 10 ^-5 M <<<<< answer

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