What Went Wrong in Calculating the Volume of Gas from Decomposition?

In summary, the conversation discusses a problem related to the decomposition of hydrogen peroxide and the calculation of the volume of oxygen gas produced. The correct answer is half of the initial calculation and it is suggested that the TA in charge of solving problem sets may have made a mistake in the answer key. The expert summarizer notes that the mole of peroxide yields half mole of oxygen gas and suggests bringing any necessary corrections to the attention of the TA or instructor.
  • #1
Pengwuino
Gold Member
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So here's the problem I am faced with... dum dum dummmm

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe):

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

What volume of pure O2(g), collected at 27 °C and 746 torr, would be generated by decomposition of 61.2 g of a 50.0 % by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

So first I figured that there will be 30.6g of hydrogen peroxide. Then I figured that there are 0.4498 moles of O2. Then using the ideal gas law...

V=nRT/P

I got ((0.4498)(0.08206)(27+273.15))/(746/760) = 11.3 Liters of O2

But supposedly I am wrong. Where did I go so horribly horribly wrong?

Oops, and i just punched in 22.6 Liters and it says I'm right... so where did I divide by 2 where I shouldn't have?
 
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  • #2
It's impossible to locate your error since you have not shown your work.
 
  • #3
Pengwuino said:
So here's the problem I am faced with... dum dum dummmm



So first I figured that there will be 30.6g of hydrogen peroxide. Then I figured that there are 0.4498 moles of O2. Then using the ideal gas law...

V=nRT/P

I got ((0.4498)(0.08206)(27+273.15))/(746/760) = 11.3 Liters of O2

But supposedly I am wrong. Where did I go so horribly horribly wrong?

You didn't read/remember the stoichiometry of the decomposition.
Oops, and i just punched in 22.6 Liters and it says I'm right... so where did I divide by 2 where I shouldn't have?

22.6? Gotta be a TA solving the problem sets --- and getting wrong answers, as is usual. You've been told where you didn't divide by two when you should have, and it's probably the same place the TA multiplied mistakenly.
 
  • #4
Tom Mattson said:
It's impossible to locate your error since you have not shown your work.

Everything was already given, all I had to do was punch in the numbers. All i needed was moles and the temperature conversion and both were done correctly. The R constant is correct as well.
 
  • #5
Bystander said:
22.6? Gotta be a TA solving the problem sets --- and getting wrong answers, as is usual. You've been told where you didn't divide by two when you should have, and it's probably the same place the TA multiplied mistakenly.

So 11.3 was correct?

I've been noticing a few of these problems are absolutely wrong lately in the homework. One problem had a very simple PV=nRT problem, EVERYTHING except 1 variable was missing and I did it and couldn't figure out the right answer. I show it to 4 other people... one about to graduate with his bs in physics, one his masters in physics, and 2 other people and no one could figure out what was wrong. The answer made sense (small volume, low pressure, normal temperature meant even smaller moles value) and it wasn't acording to the homework program.
 
  • #6
Mole of peroxide yields half mole of O2. It's called a sitting duck.

Collateral duties of TAs include solving problem sets, checking answer keys, and keeping keys up to date when instructors change the numbers from year to year to defeat the frat-rat files --- these duties are usually performed in lackluster fashion, if at all.

Your job as a student is to bring necessary corrections to TAs' and instructors' attentions in as diplomatic a fashion as possible --- if they don't take it gracefully, you've got end of term evaluation forms.
 
  • #7
Bystander said:
Your job as a student is to bring necessary corrections to TAs' and instructors' attentions in as diplomatic a fashion as possible --- if they don't take it gracefully, you've got end of term evaluation forms.

Well was I right with the first calculation (11.3)?? I am getting a little confused by everyone here :P
 
  • #8
Half mole, 11 liters, yes.
 

Related to What Went Wrong in Calculating the Volume of Gas from Decomposition?

1. What is "V of gas from decomposition"?

The "V of gas from decomposition" refers to the volume of gas produced during the decomposition of a substance. It is a measure of the amount of gas released as a result of a chemical reaction or natural process.

2. How is the "V of gas from decomposition" measured?

The "V of gas from decomposition" is typically measured using a gas syringe or a graduated cylinder. The volume of gas can also be calculated by measuring the displacement of water in a graduated cylinder.

3. What factors affect the "V of gas from decomposition"?

The "V of gas from decomposition" can be affected by various factors such as the type of substance being decomposed, the temperature and pressure conditions, and the presence of catalysts. These factors can impact the rate and extent of the decomposition reaction.

4. Why is it important to measure the "V of gas from decomposition"?

Measuring the "V of gas from decomposition" is important for understanding the chemical reactions and processes taking place. It can provide valuable information about the properties of the substance being decomposed and the efficiency of the reaction.

5. What are some common examples of "V of gas from decomposition" in everyday life?

Some common examples of "V of gas from decomposition" in everyday life include the production of oxygen and carbon dioxide during photosynthesis, the release of carbon dioxide from baking soda and vinegar, and the generation of methane gas from decomposition of organic matter in landfills.

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