How many mL of 0.024M KMn04 (Potassium Permanganate) is required to react with 20mL of 0.112M Fe2+?

Write the equation and balance it.

Calculate moles Fe^2+ = M x L
Using the coefficients in the balanced equation, convert moles Fe^2+ to moles KMnO4.
Then M KMnO4 = mols KMnO4/L KMnO4.Solve for L and convert to mL.

To determine the volume of 0.024M KMnO4 required to react with 20mL of 0.112M Fe2+, we need to use the stoichiometry of the balanced chemical equation for the reaction between KMnO4 and Fe2+.

The balanced equation is:

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

From the balanced equation, we can see that the ratio between Fe2+ and MnO4- is 5:1. This means that 5 moles of Fe2+ react with 1 mole of MnO4-.

First, let's calculate the number of moles of Fe2+ in 20mL of a 0.112M solution:

Volume (in L) = 20mL = 0.02L
Concentration (in mol/L) = 0.112M

moles of Fe2+ = concentration × volume = 0.112M × 0.02L = 0.00224 mol

Since the ratio between Fe2+ and MnO4- is 5:1, we can infer that we need 0.00224 mol of MnO4- to react with the Fe2+.

Now, let's calculate the volume of 0.024M KMnO4 needed to supply 0.00224 mol of MnO4-:

Concentration of KMnO4 (in mol/L) = 0.024M
moles of MnO4- = 0.00224 mol

Volume (in L) = moles of MnO4- / concentration = 0.00224mol / 0.024M = 0.0933 L

Finally, let's convert the volume from liters to milliliters:

Volume (in mL) = Volume (in L) × 1000 = 0.0933 L × 1000 = 93.3 mL

Therefore, 93.3 mL of 0.024M KMnO4 is required to react with 20 mL of 0.112M Fe2+.