If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN)]2+ that would be formed from 6.00 mL of 0.00180 M KSCN 5.00 mL 0.240 M Fe(NO3)3 and 14.00 mL of 0.050 M HNO3.

Question 3 options:

0.240 M

4.32 x 10^-4

0.0480

0.0460 M

Answer : The concentration of is,

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is and is excess reagent.

First we have to calculate the moles of KSCN.

Moles of KSCN = Moles of = Moles of =

Now we have to calculate the concentration of

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

Thus, the concentration of is,