Question

Suppose a solution is prepared by adding 3.00 mL (Vc) of 0.00200 M Fe3+ (Mc), 2.00 mL of 0.00200 M SCN- and 5.00 mL 0.500 M HNO3. The required volume of the diluted solution, Vd =10.00 mL.

Answer 1

The balanced chemical equation for the reaction is:

Fe3+ + SCN- -> FeSCN2+

From the equation, we see that the stoichiometry of the reaction is 1:1, which means that 1 mole of Fe3+ reacts with 1 mole of SCN- to form 1 mole of FeSCN2+.

Using the given concentrations and volumes, we can calculate the moles of each reactant:

Moles of Fe3+ = (0.00200 M) x (0.00300 L) = 6.00 x 10^-6 mol
Moles of SCN- = (0.00200 M) x (0.00200 L) = 4.00 x 10^-6 mol

To determine which reactant is the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometry of the reaction. Since the stoichiometry is 1:1, the reactant with the smaller number of moles is the limiting reagent.

In this case, SCN- has the smaller number of moles, which means that it is the limiting reagent. The Fe3+ is in excess.

The number of moles of FeSCN2+ formed is equal to the number of moles of SCN- used, which is 4.00 x 10^-6 mol.

To calculate the concentration of FeSCN2+ in the solution, we need to use the moles of FeSCN2+ formed and the total volume of the solution:

Moles of FeSCN2+ = 4.00 x 10^-6 mol
Total volume of solution = 0.0100 L

Concentration of FeSCN2+ = Moles of FeSCN2+ / Total volume of solution

Concentration of FeSCN2+ = (4.00 x 10^-6 mol) / (0.0100 L)

Concentration of FeSCN2+ = 0.000400 M

Therefore, the concentration of FeSCN2+ in the solution is 0.000400 M.