30cm^3 of a mixture of methane and ethane were mixed with 100cm^3 of oxygen at ordinary temperature and pressure and exploded. After cooling the residual gases occupied 61.5cm^3. Find the percentage by volume of each gas in the mixture, methane (CH4) and ethane(C2H6).
LC, didn't I work one just like this (different numbers as I recall for the mixture was 24 cc of CH4 + C2H5) three or four days ago? I know I did. Look at your notes or other data. This one is done the same way and it isn't all that difficult.
I thought it was similar so I wrote...
x + y = 30cm^3
x + 2y = 61.5cm^3
i.e
2x + 2y =60cm^3
x + 2y = 61.5cm^3
therefore
x= -1.5cm^3(volume of methane)
and
y= {61.5-(-1.5)}/2= 31.5cm^3(volume of ethane)
However my teacher said the volume i got for ethane is wrong, because my second equation(x+2y) should not be equal to 61.5cm^3.... so this is why I am lost. I tried to follow your previous example though. So I need your help again.
As a hint she she gave us this...
61.5cm^3- (A + 2B)= 100- (2A+ 3 and a 1/2B)...I have no idea what this is.
Well LC, you are right. This one is not so simple. This one you know is wrong when you come out with a negative volume for CH4. Give me some time to think about it but I may not get back until tomorrow. It's bed time now.
ok thanks
It would help if you posted that old problem to let me see exactly how it was worded.
The one from last week?
If yes...Here it is...
24cm^3 of a mixture of methane and ethane were exploded with 90cm^3 of oxygen. After cooling to RTP the volume of gases was noted and it decreased by 32cm^3 when treated with KOH. Calculate the composition of the mixture.
Yes, that's the one. Thanks. I just wanted to see exactly how it was worded. I assume that answer was right BECAUSE the KOH dissolved all of the CO2 so decreasing by 32 meant that was the amount of CO2 generated. This problem is a little different since we don't know how much CO2 was formed. I think the answers are 17cc for ethane and 13cc for methane but I did that by iteration.
Yes the answer to the first question was correct, that's what my teacher got too... but she did it like this. Either way its correct.
x + y = 24cm^3
x + 2y = 32cm^3
then
32 - 2y + y = 24cm^3
32-y = 24cm^3
y= 8cm^3(volume of ethene)
x + 8 = 24cm^3
x= 16cm^3(volume of methane)
Here are the equations and some explanation.
CH4 + 2O2 ==> CO2 + H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O
--------------------------
The first equation is x + y = 30
Here is how you get the second equation.
x cc CH4 gives x cc CO2
x cc CH4 USES 2x O2.
y cc C2H6 gives 2y cc CO2
y cc C2H6 USES (7/2) cc O2
We had 100 cc O2 to start.
So the total is
xcc CO2 - 2x cc O2 + 2y cc CO2 - (7/2)y cc O2 + 100 cc = 61.5 and that can be simplified to
-x cc - (3/2)y cc = 61.5-100 = -38.5
So to make it look a little better,
x + y = 30
-x - (3/2)y = -38.5
(Your teacher may not like the - equation but it works).
I solved it and obtained 17 cc for y and 13 cc for x (just like my iteration said it was).
Interesting. Isn't chemistry fun?
Lol when you explain it. Not when my teacher teaches me, I never understand a word she says. Anyways thanks alot for staying up to help me.
There's something yo should know though...I wrote the reaction like how you did.
CH4 + 2O2 ==> CO2 + H2O
and my teacher said H2O is not part of the residual gases. She said H2O should be replaced with O2.
CH4 + 2O2 ==> CO2 + O2
I guess this changes a lot of things. So I'll go to class with what you gave me and see what happens and maybe tomorrow (because its late for you)you can tell me whats your answer with the O2 instead. Thanks a lot again.
****There's something you should know though...I wrote the reaction like how you did.
(CH4+ C2H6) + 2O2 ==> CO2 + H2O
and my teacher said H2O is not part of the residual gases. She said H2O should be replaced with O2.
(CH4+ C2H6) + 2O2 ==> CO2 + O2
****
I made a typo in my last post. I wrote about half way into that second equation explanation a (7/2)cc and it should have been (7/2)y cc if you can locate and make the change.
As for the equations, for the first problem it is true that H2O was not one of the residual gases; however, we didn't relay on that because the KOH absorbed all of the CO2 and they told ushow much of the gas was CO2 and I still would write the water in the equation because that's what is formed on combustion.
For the second problem, I still like my combustion equations as is because that's what is formed on combustion. I don't think it's accurate to write a combustion equation as CH4 + 2O2 ==> CO2 +O2. In the first place it isn't balanced (with O2) and in the second place it isn't complete without the H2O. But don't get in an argument with your teacher. Your teacher is ALWAYS right.
More about the equations. As I've already said whether H2O is a residual gas in problem #1 is a non-issue. It isn't of course; but the problem never mentioned anything about residual gases.
In problem #2 I agree H2O is not part of the residual gas (and that term is used) but I stick with my argument that a combustion equation of a hydrocarbon produces CO2 and H2O. I did not use H2O as part of the residual gas when I made up equation #2 of problem #2. Thanks for your help in solving this.