Question

The amount of iron in a sample can be determined by a titration using dichromate ion,
Cr2 O 72-, in acidic solution. The iron is first converted to Fe2+ , which is then titrated by
the dichromate ion. The reaction is:
6 Fe2+ (aq) + Cr 2 O 72-(aq) + 14 H+ → 6 Fe3+ (aq) + 2 Cr 3+ (aq) + 7 H2 O
How many grams of iron are present in a sample if it required 42.7 mL of 0.0180 M
Na 2 Cr2 O 7 (aq) solution for the titration described by the redox equation above?

Answer 1

Answer-Explanation:

The balanced chemical equation for the reaction is:

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

We are given that the concentration of the Na2Cr2O7 solution is 0.0180 M and that 42.7 mL of this solution was used to titrate the iron sample. We can use this information to calculate the number of moles of Na2Cr2O7 that were used in the titration.

The number of moles of Na2Cr2O7 is equal to the concentration of the solution multiplied by the volume of the solution.

Number of moles of Na2Cr2O7 = 0.0180 M × 0.0427 L = 0.000766 mol

We can use the balanced chemical equation to determine the molar ratio of Na2Cr2O7 to Fe2+.

1 mol Na2Cr2O7 : 6 mol Fe2+

We can use this molar ratio to calculate the number of moles of Fe2+ that were present in the sample.

Number of moles of Fe2+ = 0.000766 mol Na2Cr2O7 × 6 mol Fe2+/1 mol Na2Cr2O7 = 0.000459 mol Fe2+

We can then use the molar mass of iron to calculate the mass of iron that was present in the sample.

Mass of iron = 0.000459 mol Fe2+ × 55.845 g/mol Fe2+ = 0.0258 g Fe2+

Therefore, 0.0258 g of iron were present in the sample