50cm^3 of a mixture of CO, CO2 and H2 were exploded with 25.0cm^3

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SUMMARY

The discussion centers on the calculation of the percentage composition by volume of a gas mixture consisting of carbon monoxide (CO), carbon dioxide (CO2), and hydrogen (H2) after an explosion with oxygen (O2). Initially, 50 cm³ of the gas mixture was combined with 25 cm³ of O2, resulting in a final volume of 37 cm³ at room temperature. Following treatment with aqueous potassium hydroxide (KOH), the volume decreased to 5 cm³, indicating that CO and H2 reacted with O2 while CO2 remained unreacted. The participants express confusion regarding the decrease in total gas volume and the role of water vapor in the calculations.

PREREQUISITES
  • Understanding of gas laws and stoichiometry
  • Familiarity with chemical reactions involving combustion
  • Knowledge of KOH treatment in gas analysis
  • Basic principles of gas volume measurement at room temperature and pressure (r.t.p.)
NEXT STEPS
  • Study the stoichiometry of gas reactions, particularly involving CO and O2
  • Learn about the effects of KOH on gas mixtures and its application in gas analysis
  • Explore the concept of water vapor in gas reactions and its impact on volume measurements
  • Investigate the principles of gas collection and measurement techniques at r.t.p.
USEFUL FOR

Chemistry students, educators, and professionals involved in gas analysis and reaction stoichiometry will benefit from this discussion, particularly those focusing on combustion reactions and gas treatment methods.

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Homework Statement



50cm^3 of a mixture of CO, CO2 and H2 were exploded with 25.0cm^3 of O2. After explosion, the volume measured at r.t.p. was 37.0 cm^3. After treatment with aqueous KOH, the volume was reduced to 5.0cm^3. Calculate the % composition by volume of the original mixture.



Homework Equations





The Attempt at a Solution


From this, I can safely assume that CO will react with O2 to form CO2
2CO + O2 --> 2CO2
and
2H2 + O2 --> 2H2O
and CO2 will not react with Oxygen.
When treated with KOH, the steam and carbon dioxide will all be removed. I assumed that 5.0cm^3 of oxygen is left. However, what I don't understand is how did the total volume of gas decrease from 75cm^3 (including oxygen) to 37.0cm^3?
 
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Is there any steam at r. t. p.? Water vapor? How much water vapor will there be after treatment with KOH? More? Less? The same? I assume that the KOH treatment is an aqueous solution of KOH but your instructor might have a different idea.

...However, what I don't understand is how did the total volume of gas decrease from 75cm^3 (including oxygen) to 37.0cm^3?

Three gases at the beginning. How many gases at the end (at r. t. p.)?
 


Hmm. Good question. Is there steam? If there isn't, then everything becomes complicated.
I guess the 'steam' will just stay with the KOH after treatment so that oxygen is the only gas that is left.
So, at the end, there is oxygen.
But I'm not that sure about that. Maybe not all carbon monoxide or hydrogen is reacted, so they may be collected? But, if I think it like that, I'm just complicating things.
 

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