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# Edexcel Unit 6 Past Paper Q watch

1. June 2002 Q2(b)

On heating hydrogen peroxide decomposes according to the equation:
2H2O2 --> 2H2O + O2

Hydrogen Peroxide is marketed as an aqueous solution of given 'volume strength'.
The common 20-volume solution gives 20dm3 of oxygen from 1dm3 of solution.
What is the concentration in g dm-3 of such a solution? (Molar volume of any gas at the temperature and pressure of the experiment is 24dm3)

i think that if
24dm3 gives 1 molar oxygen
then 20 dm3 gives 0.83 molar oxygen
and (as 1 mole gives 16g) 0.83 moles of oxygen equates to 13.28g
is this right?
2. I got the Mark scheme for that paper in front of me and heres the answer and ive indicated where you get the marks:

20dm3 oxygen is 20/24 = 0.833 mol (1)
Amount of peroxide in 1dm3 = 0.833 mol x 2 = 1.67 mol (1)
mass = 1.67 mol x 34 g mol-1 = 57 dm3 (1)

3. No - you need to consider moles of H2O2 in the solution not moles of oxygen.

1 dm3 of H2O2 gives 20 dm3 of O2

so 1 dm3 of H2O2 gives 20/24 moles O2 = 0.833 moles

From the 2:1 ratio in equation... moles of H2O2 in 1dm3 = 0.833 x2 = 1.66 moles

Molar mass of H2O2 = 34 ... so mass of 0.833 moles = 1.66 x 34 = 56.6 g

so there are 56.6g in 1dm3 ... concentration = 56.6 g/dm3
4. oxymoron its not 0.833 moles x 34 becuase you have 2 moles of h2o2 becuase its 2:1 ratio, so you would have to multiply 0.833 by 2.
5. (Original post by b_shaikha4321)
oxymoron its not 0.833 moles x 34 becuase you have 2 moles of h2o2 becuase its 2:1 ratio, so you would have to multiply 0.833 by 2.
Yeah I know - mis-read the original equation first time round but have already changed it

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