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

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The discussion revolves around calculating the amount of ethylene oxide produced from a mixture of ethylene and oxygen in a 1:1 molar ratio. The reaction occurs at atmospheric pressure and a corrected temperature of 556.15K. Participants clarify the use of the ideal gas law (PV=nRT) to determine the number of moles, emphasizing the importance of stoichiometry and identifying limiting reagents. The initial calculations were incorrect due to misinterpretation of gas amounts and temperature. Ultimately, understanding the mixture composition is crucial for accurate results.
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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.

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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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