Question

Can anyone please help me with the chem1 lab question?!

Answer 1

10.00 mL of 3.62E-3 M Fe(NO3)3 is mixed with 2.00 mL of 3.16E-3 M KSCN and 8.00 mL of water. The equalibrium molarity of Fe(SCN)2+ is found to be 9.78E-5 M. If the reaction proceeds as shown below, what is the value of Kc for the reaction?

Answer 2

Write a balanced chemical equation. Use an I.C.E. table to find the concentrations at equilibrium, then plug those values into it's equilibrium expression.

Answer 3

can u solve and show me steps or just show me because this is still confusing toe me

Answer 4

i can't solve it for you, but i can help you if you get stuck. Start by writing a balanced equation for the process.

Answer 5

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Answer 6

yields between scn and fescn2

Answer 7

it's not balanced, Fe3+(aq) + 2 SCN-(aq) -> FeSCN2+(aq)

now you have to find the moles with \(Molarity=\dfrac{n_{solute}}{L_{solution}}\)

now use an ICE table (initial, change, final).
Fe3+(aq) + 2 SCN-(aq) -> FeSCN2+(aq)
I
C
E

Answer 8

before you make the table, find the molarity of each after mixing (i.e. divide the moles of each by the total volume of solution, 20 mL)

Answer 9

okay molarity is - 1.22e-4 of FE, 1.86e-4 of SCN and 2.835e-6 of the total

Answer 10

okay i believe i have solved it

Answer 11

product/reacants

Answer 12

so these numbers are the molarity taking into account the total volume?

Answer 13

yes

Answer 14

okay now you need an ice table, (to figure out how much of the reactants you lost).

Answer 15

molarity is on the I colum

Answer 16

how do i approach the rest

Answer 17

so you would do this;

Fe3+(aq) + 2 SCN-(aq) -> FeSCN2+(aq)
I 1.22e-4 1.86e-4 0
C -x -2x +x
E 1.22e-4 (1.86e-4)-2x 2.835e-6

so x=2.835e-6

now find the rest of the equilibrium concentrations and plug into the eq expression

K=[prod]/[reactants]

Answer 18

the numbers in the problem have changed can u check my work

Answer 19

i made a mistake it should read

E 1.22e-4-x (1.86e-4)-2x 2.835e-6

Answer 20

post it?

Answer 21

okay heres the q 5.00 mL of 3.16E-3 M Fe(NO3)3 is mixed with 5.00 mL of 3.89E-3 M KSCN and 0.00 mL of water. The equalibrium molarity of FeSCN2+ is found to be 7.90E-5 M. If the reaction proceeds as shown below, what is the value of Kc for the reaction?

Answer 22

heres my ice chart
feno3 kscn fescn2+
i 6.32e-4 7.78e-4 0
c -x -2x +x
e6.162e-4 7.746e-4 1.58e-5

Answer 23

so product/reactant 1.58e-5/(6.162e-4)(7.746e-4)

Answer 24

so you're missing the exponent on [KSCN] from the equation

\(K=\dfrac{[Fe(SCN)_2]}{[KSCN]^2[Fe(NO_3)_3]}\)