A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeye

Question

A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeyer flask and diluted with 20 mL of 1 M aqueous sulfuric acid. To this solution is added 0.0200 M KMnO4 solution via a buret, until a pale purple color persists. This requires 22.50 mL of KMnO4 solution. What is the percent by mass of hydrogen peroxide in the original solution?

(A) 0.613%

(B) 1.53%

(C) 3.83%

(D) 7.65%

Is there any possible way to do this without writing out a balanced equation to get the ratio of moles? Its a semi-hard equation to just whip out of my pocket and wont be given on the exam. Please show all steps and thanks!

Explanation / Answer

2 MnO4^- + 5 H2O2 + 6 H^+ ---> 2 Mn^2+ + 8 H2O + 5 O2

2 mol MnO4^- = 5 mol H2O2

no of mol of KMnO4 consumed = 22.5*0.02 = 0.45 mmol

no of mol of H2O2 reacted = 0.45*5/2 = 1.125 mmol

amount of H2O2 reacted = n*M

                       = 1.125*10^-3*34

                       = 0.03825 g

% of H2O2 in sample = 0.03825/1*100

                    = 3.83%
answer: C

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