How Much Ethylene Oxide Can Be Produced from 35100 L of Ethylene and Oxygen?

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Discussion Overview

The discussion revolves around calculating the amount of ethylene oxide that can be produced from a specified volume of ethylene and oxygen, given a 1:1 molar ratio of the gases. The context is a homework problem involving stoichiometry, gas laws, and chemical reactions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the balanced chemical equation for the reaction and expresses uncertainty about the pressure value to use, suggesting 1 atm as a possible assumption.
  • Another participant confirms that atmospheric pressure is indeed 1 atm and advises on selecting the appropriate gas constant for the calculations.
  • A participant calculates the number of moles of ethylene oxide produced using the ideal gas law, but questions the correctness of their result.
  • Some participants suggest checking the initial amounts of gases and stoichiometry, indicating that the temperature conversion may have been incorrect.
  • There is a discussion about whether there is enough oxygen available for the reaction, with one participant suggesting that using the ideal gas law may not be necessary since the mixture composition is known.
  • Another participant reiterates the importance of determining the limiting reagent based on the provided molar ratio.

Areas of Agreement / Disagreement

Participants generally agree on the need to check the limiting reagent and the stoichiometry of the reaction. However, there is no consensus on the specific calculations or the correct approach to take next, as some participants express confusion about the steps involved.

Contextual Notes

Participants mention potential errors in temperature conversion and the need to clarify the initial amounts of gases. There are unresolved questions regarding the application of the ideal gas law and the implications of the 1:1 molar ratio on the limiting reactant.

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Homework Statement



Ethylene oxide is produced industrially from the reaction of ethylene with oxygen at atmospheric pressure and 283 oC, in the presence of silver catalyst.

C_{2}H_{4} + O_{2} \rightarrow C_{2}H_{4}O

Assuming 100 % yield, how many kg of ethylene oxide can be produced from 35100 L of a mixture containing ethylene and oxygen in 1:1 molar ratio?

Homework Equations





The Attempt at a Solution



First I balanced the equation like so,

2C_{2}H_{4} + O_{2} \rightarrow 2C_{2}H_{4}O

I then converted the temp. from celsius to kelvin so,

525.15K

I'm not sure what value I should use for pressure, in a question like this am I just assuming that P = 1atm?

I then applied PV= nRT and solved for moles. Once I have the number of moles I used the molar mass to get it into grams and then I converted grams to kilograms.

I still got the answer wrong however.

What am I doing wrong?

EDIT:Also, what value for the gas constant do I use that will work with the units K, mol, atm, and L?
 
Last edited:
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Show details of your work.

You are told it is happening at atmospheric pressure, and that means 1 atm.

R value - check whichever fits the units. There are many lists of http://www.chemistry-quizzes.info/faq.php on the web.
 
n = \frac{PV}{RT}

n = \frac{1atm \cdot 35100L}{0.082 \cdot 525.15K} = 815.10 \text{mols of Ethylene oxide}

Molar mass of Ethylene oxide 44.05 g/mol

So,

44.05\text{g/mol} \cdot 815.10\text{mols} = 35905.07g = 35.91kg

This answer is incorrect.

What am I doing wrong?
 
Last edited:
Think about initial amounts of gases and stoichiometry.

And check your kelvins.

--
 
Do you have enough oxygen?
 
So I should use PV=nRT to find the mols of oxygen, then I'll know whether it's a limiting reactant or not, correct?
 
Yes and no. Yes - you have to check what is a limiting reagent. No - you don't need PV=nRT for that. You are already told what is mixture composition.

--
 

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