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1. Propane, C3H8 is a hydrocarbon that is commonly used as fuel for cooking.

a)Write a balanced equation for the complete combustion of propane gas, which yields CO2(g) and H20(l)

b) Calculate the volume of air at 30 degrese Celcius and 1.00 atm needed to burn completely 10.0 grams of propane. Assume the air is 21% O2 by volume.

c)The heat of combustion of propane is -2200.1 kj/mol. Calculate the heat of formation, deltaHf, of propane given that deltaHf of H20(l) = -285.3j/gK and deltaHf oc COs(g) = -395.5kj/mol.

d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transfered to 8kg of water (specific heat = 4.184J/gk) calculate the increase in temperateure of water


2. A mixture of H2(g) and 2mL of H2O(l) are present in a .5 litre rigid container at 25 deg. C. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1146 mm Mercury. (The equilibrium vapor pressure of water is at 25 deg C us 24 mL mercury.) The mixture is sparked, and H2 and O2 react until one reactant is completely consumed.

a) identicy the reactant remaining and calculate the number of moles remaining

b) Calculate the total pressure in teh container at the conclusion of teh reaction if the final temp. is 90C. (The equilibrium vapor pressure of water at 90C is 526 mm mercury)

c) Calculate the number of moles of water present AS VAPOR in the container at 90C

By Joel_Set (Joel_Set) on Friday, October 03, 2003 - 10:21 pm: Edit

okay I haven't gone into chapter 5 yet (Gases and what not) ;) so.. i can only answer the first one..

a) C3 H8 + 5 O2--> 3C02 + 4H20
Just remember... methane always combusts with O2... hope this helped just a little

By Montydsw11 (Montydsw11) on Friday, October 03, 2003 - 11:09 pm: Edit

hehe, thanks joel... anyone else?

By Montydsw11 (Montydsw11) on Friday, October 03, 2003 - 11:10 pm: Edit

oh btw joel, you say chapter 5... what book are you using? Zumdahl?

By Joel_Set (Joel_Set) on Saturday, October 04, 2003 - 09:44 pm: Edit

yeap.. i think everyone does :/ it's sooo hard tho..

By Montydsw11 (Montydsw11) on Saturday, October 04, 2003 - 10:23 pm: Edit

Anyone else? I am really having problems with this. Please?

By Montydsw11 (Montydsw11) on Sunday, October 05, 2003 - 10:19 am: Edit

bump

By Jszab (Jszab) on Sunday, October 05, 2003 - 11:37 am: Edit

OKay. I haven't started preparing yet. But watch my time ;).
I would be pleased If you would like answer my questions. Would you?

By Jszab (Jszab) on Sunday, October 05, 2003 - 11:43 am: Edit

V = (50/44 moles * R * T)/0,21 P

By Jszab (Jszab) on Sunday, October 05, 2003 - 12:01 pm: Edit

Second is harder. So ok. We have 2200.1 from combustion. But we need about 2350 to have all on the left. So [It might be...hmm false?] we have three carbon-dioxide, and 4 molecules of water.

These compounds are 2200.1 below energy level of propan.
And we need app. 2350 j, to form enough of these substances. So, all energy that is needed to form propan is 2350 j + 2200 j.
Aaa. I will never make such tasks on SAT II Chem, 1 minut per question. Is it from there? I have one year to Chemistry...

By Jszab (Jszab) on Sunday, October 05, 2003 - 12:02 pm: Edit

I will be back in the moment. My biology olympiad work is waiting. [i must change cockroaches ;)].

By Jszab (Jszab) on Sunday, October 05, 2003 - 12:19 pm: Edit

OF course it is fake. It is obvious. Logically,
the propan has higher energy. It is wasting it during combustion. Sum of x*1+x*2 - 2200.1 kj/mol.

Or on the other hand:
We have -2350 kj.And then we are 2200.1 below.

By Jszab (Jszab) on Sunday, October 05, 2003 - 12:27 pm: Edit

[below means closer to 0]
Third one.
O that's easy, i think.

One mole of propan is weighting 44g/mol.
we have then, 30/44 moles.

30/44 moles * 2200.1 kj/mole.
1500 kj? Calculate it. I don't know. Assume that 1500.

Mass of water: 8 kilos.
app. 4,2 Kj for one kilo, per one degree.
1500(kj)/(4.2[kj/K*kg] * 8kg) = 1500 kj/33.6 = 44,64 K.

By Jszab (Jszab) on Sunday, October 05, 2003 - 01:00 pm: Edit

More tommorow.
There is more to count, because of equilibrium vapor pressure.
You have to subtract v. pressur of water.
And know, what volume will be there. You need two equals. Sorry, I have my physics.

By Montydsw11 (Montydsw11) on Sunday, October 05, 2003 - 01:15 pm: Edit

jzsab, thanks for your help. I am having a little trouble following your steps... do you think you could rewrite it with subheadings for each step of the problem? For example, the answer to part A would look like

A).. etc

Anyway, thanks for help if you cant do it thats still fine.

By Jszab (Jszab) on Sunday, October 05, 2003 - 04:09 pm: Edit

I will post something tommorow. Or, send me an sms, if you need it for monday. +48601616111
These questions are hard, and it would be definitely good for me, to make it before 1st november... Ou. I started from B, think that a) is correct. Try the web. Where did you get those questions?

By Montydsw11 (Montydsw11) on Sunday, October 05, 2003 - 05:52 pm: Edit

Thanks Jszab, I need it for monday night not monday morning, hopefully that gives you enough time.

I got those questions from an AP review packet. Thanks again I am looking forward to your detailed explanations!

By Nhlgoalie (Nhlgoalie) on Monday, October 06, 2003 - 01:34 am: Edit

A)C3H8 + 5 O2--> 3C02 + 4H20

B)10 g Propane = (10)/(44) mol Propane

Using Part A:

Combusting 10/44 mol Propane uses 50/44 mol oxygen gas

Using PV=nRT

(1 atm)(V)=(50/44 mol)(303 K)(.0821 l-atm/mol-K)

V=28.27 liters

C)3(-395.5)+4(-285.3) = -2327.7 kJ

This is the heat of formation of the entire right side of the equation

Now we are told that when you combust propane the reaction gives off 2200 kJ of heat. Thus the difference between the heat of formation and the heat of combustion must be the heat of formation of the right side.

2327 kJ – 2200 kJ = 127 kJ

Since the heat of formation of oxygen gas is zero (since gaseous diatomic oxygen is oxygen’s standard state), the heat of formation for propane must be 127 kJ. I’m guessing the sign would be negative (ie. –127 kJ), but I took chemistry a while back, so I may be wrong.

It's really late and I'm really tired, so I'm assuming Jszabs answer for D is right, but I didn't look at it. Email me if it isn't right and you need help. I'm gonna look at your second question, but I'm quite tired, so we'll see if I do it now or not.

By Nhlgoalie (Nhlgoalie) on Monday, October 06, 2003 - 01:52 am: Edit

Number 2

I can do questions like this all the time, but I’m struggling with the actual question itself. It is horribly phrased and difficult to understand. I wouldn’t bother with it if I were you, but part A should be something like this…

A) PV=nRT

Subtract out waters vapor pressure and plug in

(1122/760 atm)(.5 L)=n(.0821 l-atm/mol-K)(298 K)

n = .030 mol (note: I don’t have a calculator, so I’m sorry for any bad math)

So you have .015 moles of diatomic hydrogen and .015 moles of diatomic oxygen

Now the question is REALLY badly written, but I’m guessing you’re supposed to examine:

2 H2 + O2 à 2 H20 (l)

In this case you would have oxygen remaining and you’d have .0075 moles left (using stoichiometry)