s asked in Science & MathematicsChemistry · 1 month ago

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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  • 1 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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