Hutchoo's Unit 2 Chem questions.

[Hutchoo]

Hey guys, I'm a bit confused with this stoich stuff! Hopefully you can all help me.

Anyway, the question is from chapter 15, question 4 from Heinemann 1/2, but I'll type it out.

Octane is a component of petrol. It burns in oxygen to produce carbon dioxide and water. Energy is released during this reaction. The equation for this reaction is:

a. Calculate the mass of oxygen required to react with 200g of octane.
b. Calculate the mass of carbon dioxide produced in part a.

For (A) All I can understand and do is figure out the ratio and the mole of Octane..
The ratio being 25/2 and the mole of octane being 1.75M
Where do I go on from here?

Much appreciated.

[b^3]

Just note it's 1.75 mol not 1.75 M as it is amount not concentration.
So we know the ratio to be 2 mol of octane to 25 mol of oxygen
so the amount of oxygen needed for 1.75 mol of octane will be
=25/2*1.75
=21.93mol
Find mass = 21.93*32
=701.7g

now for b)
so the ratio of octane to CO2 is 2:16 or 1:8
so we are saying that 1 mol of octane will produce 8 mols of CO2
so the n(CO2)=8/1 * 1.75
=14.04mol
so the mass of CO2 produced = 14.04 *M
=14.04*44
=617.5g

Just remember it's all about ratios.
From the ratio you determine the amout you get out for what you put in.
I hope the calculations are right.

[Hutchoo]

Thanks a lot for that

ZZZ. I understand the process of questions very similar to that one you just did, but ones like these still confuse me sigh.


Calculate the masses of potassium perchlorate (KClO₄) and potassium chloride (KCl) produced from 26.0 g of potassium chlorate (KClO₃) reacting according to the equation:

4KClO₃(s) 3KClO₄(s) + KCl(s)

[REBORN]

This is fundamental. Perfect this now before you hit [u3].

n(KlClO₃) = 26/molar mass of it.

n(KlClO₄) = 3/4 x n(KlClO₃)

Then use m = n x M

[Hutchoo]

This is fundamental. Perfect this now before you hit [u3].

n(KlClO₃) = 26/molar mass of it.

n(KlClO₄) = 3/4 x n(KlClO₃)

Then use m = n x M

lol. So simple. Thanks!

[Lasercookie]

I'm guessing figuring out the mole ratios is where you get stuck (well that's what confused me initially anyway).
Sorry if you're already on top of this stuff. I guess it won't hurt to revise this stuff.

Just some stuff on mole ratios:
http://dbooth.net/mhs/chem/moles.html
http://www.wisc-online.com/objects/ViewObject.aspx?ID=GCH7304 (interactive flash thing, kind of like the stuff you find on textbook cds)
http://www.khanacademy.org/video/stoichiometry?playlist=Chemistry (I will definitely have to donate a decent amount to KhanAcademy later in life lol)

edit: I think I need to stop overusing the word 'stuff'

[Hutchoo]

I'm guessing figuring out the mole ratios is where you get stuck (well that's what confused me initially anyway).
Sorry if you're already on top of this stuff. I guess it won't hurt to revise this stuff.

Just some stuff on mole ratios:
http://dbooth.net/mhs/chem/moles.html
http://www.wisc-online.com/objects/ViewObject.aspx?ID=GCH7304 (interactive flash thing, kind of like the stuff you find on textbook cds)
http://www.khanacademy.org/video/stoichiometry?playlist=Chemistry (I will definitely have to donate a decent amount to KhanAcademy later in life lol)

edit: I think I need to stop overusing the word 'stuff'

<3
Thank you so much, mate. I'll definitely go over them today (had a quick look and it looks really good!)