Asked by damon
From a hydrogen peroxide (diluted in water) a sample of 1.00 g is removed, acidifing it with H2SO4 and then titrated with a solution of 0.20 M of KMnO4, requiring 17.6 mL of KMnOH. What is the percentage of H2O2 contained in the bottle?
Answers
Answered by
DrBob222
5H2O2 + 2KMnO4 + 3H2SO4 = 5O2 + 2MnSO4 + K2SO4 + 8H2O
millimols KMnO4 used = mL x M = 17.6 mL x 0.2 M = ?
Convert to mmols H2O2 = ? mmols KMnO4 x (5 mols H2O2/2 mols KMnO4) = ??
Change mmols H2O2 to mols (divide by 1000)
grams H2O2 = mols H2O2 x molar mass H2O2 = ?
%H2O2 = (grams H2O2/1.00 g sample)100 = ?
NOTE: This is the %H2O2 in the diluted sample. It is not the %H2O2 in the bottle, before dilution, and without know the dilution factor one may not calculate how much is in the bottle of H2O2.
millimols KMnO4 used = mL x M = 17.6 mL x 0.2 M = ?
Convert to mmols H2O2 = ? mmols KMnO4 x (5 mols H2O2/2 mols KMnO4) = ??
Change mmols H2O2 to mols (divide by 1000)
grams H2O2 = mols H2O2 x molar mass H2O2 = ?
%H2O2 = (grams H2O2/1.00 g sample)100 = ?
NOTE: This is the %H2O2 in the diluted sample. It is not the %H2O2 in the bottle, before dilution, and without know the dilution factor one may not calculate how much is in the bottle of H2O2.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.