Write the balanced equation for the reaction between hydrogen peroxide and potas
ID: 891386 • Letter: W
Question
Write the balanced equation for the reaction between hydrogen peroxide and potassium permanganate that occurs in acidic solution. H2O2+KMnO4 rightarrow What is the molar ratio of reactants H2O2 to KMnO4 in this titration? The moles of KMnO4 reacted in each trial can be calculated using the volume (liters) of the KMnO4 added and the molarity (i.e. - moles/liter) given for the permanganate solution Calculate the moles of KMnO4 reacted in each trial and fill these value two the Data and Results Table. The molar ratio of the reactant as expressed in the balanced equation, can now used to calculate the moles of hydrogen peroxide reacted in each trial. Show the calculation, of the moles of H2O2, and fill the values into the Data and Results table. How many grams of H2O2 were present in each trial? All of these grams of hydrogen Peroxide were present in the original H2O2 sample that you weighed out. Knowing the mass of the original commercial sample of hydrogen Peroxide and having calculated grams of hydrogen Peroxide in this original sample using your titration data, calculate the percent hydrogen Peroxide by mass in each original sample. Shoe the set up of the calculation below
Explanation / Answer
1. the balanced equation here would be,
2KMnO4 + 6H+ + 5H2O2 ---> 2MnO2 + 8H2O + 5O2
2. the molar ratio of H2O2 to KMnO2 is 2.5 : 1
3. moles of KMnO4 reacted = molarity x volume
For trial 1.
moles of KMnO4 reacted = 0.111 M x 31.95 mL = 3.55 mmoles
For trial 2
moles of KMnO4 reacted = 0.111 M x 26 mL = 2.89 mmoles
4. For trial 1.
moles of H2O2 reacted = 2.5 x 3.55 mmoles = 8.88 mmoles
For trial 2.
moles of H2O2 reacted = 2.5 x 2.89 mmoles = 7.23 mmoles
5. For trial 1
grams of H2O2 = moles x molar mass = 8.88 mmoles x 34.015 g/mol = 0.302 g
For trial 2.
grams of H2O2 = 7.23 mmoles x 34.015 = 0.246 g
6. For trial 1
% H2O2 by mass = 0.302/5.959 x 100 = 5.07%
for trial 2
% H2O2 by mass = 0.246/6.256 x 100 = 3.93%
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