Hercules Chem questions :D

[HERculina]

A 2L sample of gaseous hydrocarbon is burnt in excess oxygen. The only products of the reaction are 8L of CO2 (g) and 10L of H20 (g), all at 100 degrees Celsius and 1atm pressure.
The formula of the hydrocarbon is
A.CH
B.C3H4
C.C4H10
D.CSH10

does the fact that theyre all gases have something to do with it? volume-volume stoich? or nah? :/

[b^3]

PV=nRT
Now temperature and pressure stay constant.
So kV=nR (k and r are just propontionality constants, so we can replace them with a single constant, basically V=aN)
i.e. V is proportional to n
Now n(C)=n(CO₂)=8 mol
n(H)=2n(H₂O)=20 mol

                       C                H
n                     8                20
                       8/8=1          20/8=2.5
ratio                  2                   5

So empirical formula = C₂H₅

Now the only answer that is a multiple of this is C. C₄H₁₀
Back-checking
(other working if you want to do the rest the long way, this part isn't nessecary but may help in the understanding, altough now that I look at it, is kinda pointless. Anyway) It is undergoing combustion so the basic equation is
aC_xH_y+bO₂--->cCO₂+dH₂O
now c:d=8:10=4:5
and C_xH_y=C₂H₅
So we now have
aC₂H₅+bO₂--->4CO₂+5H₂O
balance it
2C₂H₅+6.5O₂--->4CO₂+5H₂O
now since 2C₂H₅ is not an answer but C₄H₁₀ is the equation becomes
2C₄H₁₀+13O₂--->8CO₂+10H₂O

EDIT: also note that 2 mol of the hydrocarbon is producing 8 mol of CO₂ and 10 mol of H₂O as stated in the question.

EDIT2: fixed balancing mistake

[HERculina]

wow thanks b^3 that really helped me

[Deank]

PV=nRT
Now temperature and pressure stay constant.
So kV=nR (k and r are just propontionality constants, so we can replace them with a single constant, basically V=aN)
i.e. V is proportional to n

I'm confused at that part. lol

[HERculina]

you know for the 2nd method, when you balance why is there 18 oxygen on LHS but 13 oxygen on RHS

[b^3]

PV=nRT
Now temperature and pressure stay constant.
So kV=nR (k and r are just propontionality constants, so we can replace them with a single constant, basically V=aN)
i.e. V is proportional to n

I'm confused at that part. lol

Just saying that since PV=nRT, the pressure is staying constant at 1 atm and the temperature is constant at 100 C.
So 101.3V=n*8.31*100
the 8.31*100/101.3 can just be represented by a proportionality constant. I.e. V=kn, as v increases n increases and vice-versa.

you know for the 2nd method, when you balance why is there 18 oxygen on LHS but 13 oxygen on RHS

Sorry little mistake, I'll fix it now. And note that it isn't a second method, it was just to prove to myself that it was right (I was just back checking).

[HERculina]

could't it be a second method though? like after all the balancing, you go wat under C would times two to equal 8 and then what under H would times 2 to equal 20

[Deank]

Thanks for the clearing up b^3

[b^3]

could't it be a second method though? like after all the balancing, you go wat under C would times two to equal 8 and then what under H would times 2 to equal 20

I don't want to exactly say yes but (I think) you still need to be showing the deriving the empirical formula bit (although I'm not 100% sure).

[HERculina]

ohh ok i just thought that might have worked too cause V is directly proportional to n? but idunno if thats accurate or not too
BTW what textbook did you use b^3 in year 12 ?

[b^3]

ohh ok i just thought that might have worked too cause V is directly proportional to n? but idunno if thats accurate or not too
BTW what textbook did you use b^3 in year 12 ?

Heinemann (not enhanced, not that it makes much difference if its enhanced or not)

[HERculina]

OHH OK. It looks like Heinemann has a lot of chapters to go through tho and it doesn't reallly have a lot of questions for each chapter

[funkyducky]

Just to offer a different perspective, I would have done:

1. Since pressure and temp are both constant, the mole ratio is the same as the volume ratio (as b^3 showed), or. 2:8:10 or 1:4:5
2. Consider: 1 CxHy + ? O2 -> 4 CO2 + 5H2O
Where CxHy is a hydrocarbon.
By equating the C, H and O on both sides, we have x=4, y=10, ie. C4H10 is the answer. Of course, in my head, I didn't use x and y, I just thought "you end up with 4c, 10H, 13O,so you must have started with that much"

Anyway, the answer is not important - what,you should learn from this question is that mole ratio=volume ratio at constant temperature and pressure, as b^3 derived.

[HERculina]

Ahhkays i seee another q. When do you know if reactants are present in their stoichiometric ratio?

[b^3]

Ahhkays i seee another q. When do you know if reactants are present in their stoichiometric ratio?

Calculate the mol of each of the reactants. Then using the balanced equation you should be able to work out how much of the second reactant is needed if there is so much of the first reactant present. If the value comes out the same as what you have then they are in the right ratio and nothing is the limiting reagent and nothing is in excess.

e.g. You are given 1 mol of Na and 1 mol of Cl₂
So write out a balanced equation
2Na(s)+Cl₂(g) ---> 2NaCl(s)

Now for every 2 mols of Na, 1 mol of Cl₂ will be consumed.
i.e. ratio 2:1
So 1 mol of Na will consume 1/2=0.5 mol of Cl₂, so in this case since it is not 1 mol of Cl₂ they are not in their stoich ratio. i.e. Cl₂ is in excess and Na is the limiting reagent.