Chemistry - Stoichiometry. Quantitative Chemistry Question

[Bishup]

Hey guys I'd really appreciate it if someone could help me with this question.

Question 1

100 cm 3 of ethene, C2 H 4 , is burned in 400 cm3 of oxygen, producing carbon dioxide and some liquid

2.

water. Some oxygen remains unreacted.

(a) Write the equation for the complete combustion of ethene.

(b) Calculate the volume of carbon dioxide produced and the volume of oxygen remaining.

Question 2

The combustion of 45.92 grams of a compound which contains only C, H and S yields 70.77 grams of CO2 and 7.24 grams of H2O. What is the empirical formula of the compound.

[Sandwich]

Question 1

100 cm 3 of ethene, C2 H 4 , is burned in 400 cm3 of oxygen, producing carbon dioxide and some liquid

water. Some oxygen remains unreacted.

(a) Write the equation for the complete combustion of ethene.

C2H4 + 302 ---> 2CO2 + 2H2O

(This is literally just balancing the complete combustion of ethene -- you should know what the complete combustion of an alkene/alkane looks like by now, hopefully!)

(b) Calculate the volume of carbon dioxide produced and the volume of oxygen remaining.

Okay so the basic principle they're trying to see if you understand here is the idea of a limiting factor. Basically this is that in order to make the product X you need to have Y mols of one thing and Z mols of another, so that Y+Z = X. So for every Z you need to also have Y in order to make X!

In the equation this translates to: you have to have at least 3mols of O2 to go with every single mol of C2H4. If you've not got enough of one or the other, it'll just sit there in excess as it has nothing to join with.

Sooo... you have 100cm3 of ethene, 400cm3 of oxygen and you know that the ratio is 1mol ethene : 3 mols O2. What you have to do (you can do this bit ) is work out how many mols of ethene you have in 100cm3, how many mols of O2 you have in 400cm3 and you will be able to see you have wayyy more mols of O2 than of ethene (as they've in fact already told you). Then you can say, well, for every mol of ethene which was reacted, I made 2 mols of CO2. So the number of mols of CO2 should be twice the number of mols of ethene! Then once you know how many mols of CO2 you have, all you have to do is work out what that is in terms of volume.

As for the volume of O2 remaining, you know how many mols of ethene there were, so you can assume that for every mol of ethene, 3 mols of O2 went missing because it reacted with the ethene to make the products. However, once you run out of ethene, all the rest of those mols of O2 have nothing to do and are in excess. So you take the total number of mols of O2 in 400cm3 and you subtract from it the number of mols which have been used up in the reaction (which will be 3 x the number of mols of ethene). Then you have the number of mols remaining, which you can convert into a volume.

Don't have time to try the second part, but hopefully that explains the first bit. Once you get the hang of the limiting factor/reagent, it's kinda straight forward to do these Qs.

[sweetnsimple786]

Um so I haven't done this in a bit. And my handwriting sucks. And I probably didn't explain it as well as I should have, but I hope this helps.

Edit: And yeah I forgot about the oxygen part, but Sandwich explained it way more concisely than I ever could have so yeah =) You could go through the grueling dimensional analysis [the multiplying things together], but you don't need to for this one.

Edited November 23, 2009 by sweetnsimple786

[Bishup]

I didn't mean to put the balancing question up. I could do that, if not I'd be damn worried. And thanks for the help Alice, you made it so much clearer. Thank you so much.

[sweetnsimple786]

So I wasn't able to "use full editor" on my last post?

Here's how I did #2. Hopefully it's not woefully wrong haha