in acidic solution, the dichromate ion, Cr2O7^2 - (aq) will oxidize Fe^2 + to Fe^3 + and form Cr^3+. What volume of 0.100M Cr2O7^2 - (aq) is required to oxidize 60.0mL of 0.250 M Fe^2 + (aq) ?

The concentration of Fe²⁺ is given as 0.250 M.

The volume of Fe²⁺ is given as 60.0 ml.

The concentration of Cr₂O₇²⁻ is given as 0.100 M.

The following is the reaction:

Cr₂O₇²⁻ (aq) + 6Fe²⁺ (aq) + 14H⁺ (aq) → 2Cr³⁺ (aq) + 6Fe³⁺ (aq) + 7H₂O (l)

1 mole of Cr₂O₇ reacts with 6 moles of Fe²⁺

The moles of Fe²⁺ present in the solution is:

Moles = Molarity * Volume

= 0.250 M * 60 ml * 1L / 1000 ml

= 0.015 ml

Thus, Fe²⁺ is producing 0.015 mol in the given reaction.

The volume of Cr₂O₇²⁻ required is:

1 mole of Cr₂O₇²⁻ = 6 moles of Fe²⁺

Molarity of Cr₂O₇²⁻ * volume of Cr₂O₇²⁻ - = 1/6 moles of Fe²⁺ present in the reaction

Volume of Cr₂O₇²⁻ = 0.015 mol/0.600 mol/L

= 0.025L * 1000ml / 1L

= 25 ml

Hence, the volume of Cr₂O₇²⁻ needed is 25 ml.