Question

Balance the equation and calculate the number of mmol and mg of potassium dichromate required for the oxidation of 1.7 mmol of indene. Do not forget to include hydronium ion as a reactant and water as a product. How much potassium dicrhormate (mg and mmol) corresponds to a 10 mol % excess?

Answer 1

Half reactions are:

Multiply dichromate half reaction by 4 and Indene half reaction by 3. Then add

With the balanced equation we can make the calculations for the problem. 3 mmol of indene needs 4 mmol of potassium dichromate so 1.7 mmol would need:

mmol K₂Cr₂O₇ = 1.7 mmol x 4/3 = 2.27 mmol (10% is 0.227 mmol)

MW of K₂Cr₂O₇ is = 294.2 mg/mmol

mg of K₂Cr₂O₇ = 2.27 mmol x 294.2 mg/mmol = 666.8 mg (10% is 66.68 mg)

Add 10% excess: mmol K₂Cr₂O₇ =2.5 mmol and mg of K₂Cr₂O₇ = 733.5 mg