Question #7a15c

2 Answers
Apr 16, 2015

Yes, the number of moles of oxygen gas produced by your reaction under those conditions for pressure and temperature will be 0.0025.

Hydrogen peroxide, #H_2O_2#, decomposes to give water and oxygen gas according to the balanced chemical equation

#2H_2O_(2(aq)) -> 2H_2O_((l)) + O_(2(g))#

You've collected 0.061 L of oxygen gas at 295.15 K and 1 atm, so you've got all the data you need to calculate the number of moles of oxygen gas produced by using the ideal gas law equation

#PV = nRT => n = (PV)/(RT)#

#n_(O_2) = (1cancel("atm") * 0.061cancel("L"))/(0.082(cancel("L") * cancel("atm"))/("mol" * cancel("K"))) = "0.0025 moles"#

So, if this was your first question, then yes, your reaction produced 0.0025 moles of oxygen gas.

I find the second part of your question to be a little confusing. You were given the density of the hydrogen peroxide solution, so are you supposed to use that to determine the theoretical number of moles of oxygen for this reaction?

I'm not sure what #"100%" H_2O_2 = "1.02 g/mL "# means, do you have a certain volume of hydrogen peroxide solution?

SIDE NOTE According to the additional information posted by Heather, it turns out that the initial hydrogen peroxide solution had a volume of 5 mL.

Even with the volume of the initial solution, you'd need its percent concentration to try and determine exactly how many moles you had present before the reaction.

Once you know how many moles of hydrogen peroxide you had, assume that all of the react and use the #2:1# mole ratio that exists between #H_2O_2# and #O_2# to get the number of moles of oxygen your reaction could have produced.

Apr 17, 2015

I apologise for posting another answer, but I don't want to make the first answer too long. Plus, I'm not sure this is what you're looking for, but here's what I think took place in your experiment.

So, you that your hydrogen peroxide solution had a volume of 5 mL. The density of this solution was listed at #"1.02 g/mL"#.

The weight by volume percent concentration, #"%w/v"#, for a hydrogen peroxide solution that has that density is listed at 6.6%.


This means that, for every 100 mL of #H_2O_2# solution, you get 6.6 g of hydrogen peroxide.

Your 5 mL will thus contain

#5cancel("mL") * ("6.6 g "H_2O_2)/(100cancel("mL")) = "0.33 g"#

Take a look at the balanced chemical equation for this reaction again

#color(red)(2)H_2O_(2(aq)) -> 2H_2O_((l)) + O_(2(g))#

Notice that you have a #color(red)(2):1# mole ratio between hydrogen peroxide and oxygen gas; this means that you can use the number of moles of oxygen gas your reaction produced to determine how many moles of #H_2O_2# reacted.

#0.0025cancel("moles "O_2) * (color(red)(2)" moles "H_2O_2)/(1cancel("mole "O_2)) = "0.005 moles "# #H_2O_2#

Use hydrogen peroxide's molar mass to determine how many grams would contain this many moles

#0.005cancel("moles") * "34.015 g"/(1cancel("mole")) = "0.17 g "# #H_2O_2#

Notice that approximately half of the hydrogen peroxide present in the 5-mL slution did not react. In fact, only a little over 50% of the #H_2O_2# underwent decomposition

#0.17/0.33 * 100 = "51.5%"#

The rest of the hydrogen peroxide did not take part in the reaction.

Even without doing any calculations, you can predict that the theoretical yield of oxygen gas will be approximately twice as many moles produced, since nearly twice as many moles of #H_2O_2# would react.

Assuming thet 100% of the hydrogen peroxide reacted, you'd get

#0.33cancel("g") * ("1 mole "H_2O_2)/(34.015cancel("g")) = "0.0097 moles "# #H_2O_2#

This means that you'd get

#0.0097cancel("moles "H_2O_2) * ("1 mole "O_2)/(2cancel("moles "H_2O_2)) = "0.00485 moles "# #O_2#

Rounded to two sig figs, the theoretical yield of oxygen gas will be

#n_(O_2"theoretical") = color(green)("0.0049 moles "O_2)#

Which is very, very close to being twice as big as the actual yield, 0.0025 moles.