1. H2O2(aq) --> H2O(l) + O2(g)

so I balanced it...

2H2O2(aq) --> 2H2O(l) + 1O2(g)

then it askes " Calculate the number of moles of oxygen gas produced from the complete catalyzed decomposition of 5.20ml sample of a 3.5% solution of H2O2. The density of the 3.5% solution of H2O2 is 1.01 g/ml. "

I find the mass in grams and get the moles and multiply by 3.5% and then use the molar ratios, but I am not getting the correct answers?

any Ideas? Thanks

-Chris

2.

3. regrettably I don't have an answer for you, sorry. What class is this in? I'll be doing something similar on Monday in AP Bio.

4. Chem 111

5. I think that the trick to that question is to realise that of the 5.20ml H2O2 only 3.5% (0.182ml) is actually hydrogen peroxide.

From there it should just be simple stoich.

6. I very much doubt that the 3.5% refers to volume. It would be most likely referred to weight, hence the necessity of the density.

Best regards,

César

7. Assuming the percent H2O2 is mass/volume, then the equation should be able to be set up as

5.20mL H2O2(aq) x 3.5g H2O2/100mL H2O2 x 1 mole H2O2/18 g H2O2 x

1mole O2/2 moles H2O2 = moles O2 produced

Does anyone see any problem with the setup?

Cheers,
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8. Does anyone see any problem with the setup?
Yes.
A) 3.5% is an adimensional number, i.e., has no units as it is the result of of a division between values with the same units.
B) You make no use of the density.

From my point of view it should be like follows:

5.20mL · 1.01g/mL = 5.25g of solution
5.25g · 0.035 = 0.184g of H2O2
0.184g / 18g/molH2O2 = 0.010 molH2O2
0.010 mol of H2O2 represents 0.005 mol of O2

Best regards,

César

9. The above answer is correct except for the 18g thats for H2O ... H2O2 is 34

10. Ooops!

Thank you!

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