Calculating Mass Lost in Decomposition of Hydrogen Peroxide

[nesan]

Someone check my answer please. :)

Homework Statement

The rate of decomposition of hydrogen peroxide was studied at a particular temperature. H2O2(aq) = H2O (l) + 1/2O2(g)

The initial concentration of hydrogen
peroxide was 0.200 mol/L. 10.0 s later,
it was measured to be 0.196 mol/L.

b) 0.500 L of hydrogen peroxide solution was
used for the experiment. What mass was lost
as O2 bubbled out of solution in this initial
10.0 s interval?

The Attempt at a Solution

What I did was.

0.500 L * 0.200 mol/L = 0.1

0.500 L * 0.196 mol/L = 0.098

0.1 - 0.098 = 0.002mol

0.002 * 16 = 0.032 g

What I'm doubting is.

0.500 L * 0.196 mol/L = 0.098

Won't the volume change after 10 seconds?

 

[Borek]

Yes, it will change, no, you don't have to worry - change will be negligible. You are replacing hydrogen peroxide with equimolar amount of water.

 

[sjb-2812]

http://en.wikipedia.org/w/index.php?title=Special:Cite&page=Hydrogen_peroxide&id=489569020 suggests that the density of hydrogen peroxide is significantly different to that of water; but as there is not much decomposition, this may be irrelevant.

 

[Borek]

http://en.wikipedia.org/w/index.php?title=Special:Cite&page=Hydrogen_peroxide&id=489569020 suggests that the density of hydrogen peroxide is significantly different to that of water; but as there is not much decomposition, this may be irrelevant.

Something else makes it completely irrelevant. He started with a 0.2M solution. Using density tables (or concentration calculator) you can easily check initial density to be around 1.0012 g/mL - so even after complete decomposition density change is just around 0.1%.

 

[DeeNos]

The volume may change slightly due to the release of gas, but since the concentration is given in moles per liter, the change in volume would not significantly affect the calculation. Your solution seems correct, as you have correctly used the initial and final concentrations to determine the change in moles of H2O2, and then converted it to grams of O2. However, it would be helpful to show the units in your calculation to make it clearer. Overall, your approach and solution seem to be correct.