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

[jegues]

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?

 

[Borek]

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.

 

[jegues]

$$n = \frac{PV}{RT}$$

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

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?

 

[Borek]

Think about initial amounts of gases and stoichiometry.

And check your kelvins.

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[jegues]

Think about initial amounts of gases and stoichiometry.

And check your kelvins.

--
www.titrations.info, www.chemistry-quizzes.info, www.ph-meter.info

Oh, the temperature should be 556.15K, my bad!

Can I get another nudge? It's still not obvious to me what to do next.

 

[Borek]

Do you have enough oxygen?

 

[jegues]

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?

 

[Borek]

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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[jegues]

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.

--
www.titrations.info, www.chemistry-quizzes.info, www.ph-meter.info

It says a 1:1 molar ratio, right?