1. A student mixes 5.00 mL 2.00 X 10~3 M Fe(NO3)3 with 5.00 mL 2.00 X 10-3 M KSC
ID: 625510 • Letter: 1
Question
1. A student mixes 5.00 mL 2.00 X 10~3 M Fe(NO3)3 with 5.00 mL 2.00 X 10-3 M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2+ is 1.20 X 10-4 M. Find Kc for the reaction Fe3+(aq) + SCN-(aq) (arrows) FeSCN2 (aq). Step 1. Find the number of moles Fe3+ and SCN- initially present. (Use Eq. 3.) ________moles Fe3+; ________moles SCN- Step 2. How many moles 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"1"? . __________moles Fe3+; _________moles SCN- Step 3. How many moles of Fe3+ and SCN- remain hi the solution at equilibrium? (Use Eq. 4 and the results of Steps 1 and 2.) _______molesFe3+ ________molesSCN- Step 4. What are 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 Step 5. What is the value of Kc for the reaction? (Use Eq. 2 and the results of Step 4.) Kc=_________Explanation / Answer
use the concentration and volume values they've given you to find the moles of Fe(NO3)3 andKSCNthat were initially present (before they were mixed)
moles = concentration in M x volume in L
n(Fe(NO3)3) = (2.00 x 10^-3) x (5/1000) = 0.00001 mol = 10^-5 mol
n(KSCN) = (2.00 x 10^-3) x (4/1000) = 0.000008 mol = 8 x 10^-6 mol
n(Fe(NO3)3) = n(Fe3+) = 10^-5 mol
n(KSCN) = n(SCN-) = 8 x 10^-6 mol
so n(Fe3+) initially present = 10^-5 mol
n(SCN-) initially present = 8 x 10^-6 mol
Fe3+ (aq) + SCN- (aq) <----> FeSCN2+ (aq)
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