I am working on this lab but can\'t seem to figure out how to finish it. 1. For

Question

I am working on this lab but can't seem to figure out how to finish it.

1. For the NEW bottle of hydrogen peroxide, record and calculate the following:

  

a
Volume of potassium permanganate used in each titration (mL)
50mL
b
The volume of potassium permanganate required to titrate 10 mL of the new hydrogen peroxide (mL) 18.04mL
c
How many moles of potassium permanganate were required to titrate 10mL of the new hydrogen peroxide (moles)
0.013mol
d
Given the stoichiometry of the reaction, how many moles of hydrogen peroxide were in the 10mL of the solution (moles)

e
How many grams of H2O2 were in the hydrogen peroxide solution (MW of H2O2 is 34.01) (grams)

f
1 mL of water weighs 1 g.  Using this information and the number of grams of H2O2 produced in the reaction, calculate the percent mass of H2O2 in the NEW solution.  (% mass = (mass H2O2 / mass water) * 100%)

Explanation / Answer

2MnO4- + 6H+ + 5H2O2 = > 2Mn2+ + 8H2O + 5O2

so 2 moles of potassium permanganate required for 5 moles of hydrogen peroxide

1)

c)moles of potassium permanganate were required = 18.04*0.2*10^-3 = 0.003608

d) moles of hydrogen peroxide = 0.003608/2)*5 = 0.00902 moles

e)grams of H2O2 were in the hydrogen peroxide solution =0.00902*34.01 = 0.30677 gms

f)weight of water = 10*1 = 10 gms

wt of H2O2 = 0.30677 gms

percent mass of H2O2 in the NEW solution = 100* 0.30677/(0.30677+10) = 2.976 %

2)

c)moles of potassium permanganate were required = 12.6*0.2*10^-3 = 0.00252

d) moles of hydrogen peroxide = 0.00252/2)*5 = 0.0063 moles

e)grams of H2O2 were in the hydrogen peroxide solution = 0.0063*34.01 = 0.214263 gms

f)weight of water = 10*1 = 10 gms

wt of H2O2 = 0.214263 gms

percent mass of H2O2 in the NEW solution = 100* 0.214263/( 0.214263+10) = 2.0976 %

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