chem A student mixes 5.00 mL 2.00 times 10-3 M Fe(NO3)3 with 3.00 mL 2.00 times

Question

chem

A student mixes 5.00 mL 2.00 times 10-3 M Fe(NO3)3 with 3.00 mL 2.00 times 10-3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2 + is 1.28 times 10-4 M. Find Kc for the reaction Fe3+(aq) + SCN-(aq) FeSCN2+(aq). Find the number of moles Fe3+ and SCN- initially present. (Use Eq. 3.) moles Fe3+; moles SCN- How many moles of FeSCN2+ are in the mixture at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3.) mL; moles FeSCN2 + How many moles of Fe3+ and SCN- are used up in making the FeSCN2-? moles Fe3+; moles SCN- How many moles of Fe3+ and .SCN- remain in the solution at equilibrium? (Use Eq. 4 and the results of Steps l and 2.) moles Fe3+; moles SCN- What arc the concentrations of Fe3+, SCN-, and FeSCN2+ at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3 and the results of Step 3.) [Fe3+] = M; [SCN-] = M; [FeSCN2+] = M mL What is the value of Kc for the reaction? (Use Eq. 2 and the results of Step 4.) Kc =

Explanation / Answer

2) total volume = 5+3 =8mL

hence no. of moles = 1.28 *10^-4*8*10^-3

    => no of moles = 1.02* 10^-6

And no. of moles of Fe that are used up = 1.02*10^-6

      no. of moles of SCN = 1.02*10^-6

3)the remaining moles of Fe = (10*-5) - 1.02*10^-6 = 8.98* 10^-6

and remaining moles of SCN = (6*10^-6) - 1.02*10^-6 = 4.98 *10^-6

4) concentration of Fe = 8.98*10-6 / 8*10^-3 = 1.125 *10^-3 M

    concentration of SCN = 4.98 *10^-6 /8 *10^-3 = 6.225 * 10^-4 M

concentration of FeSCN = 1.28*10^-4 M

5) Kc = 1.28*10^-4 /(1.125*10^-3 *6.225*10^-4)

   => Kc = Kc = 1.827 *10^2

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