# Chemistry multiple choice question

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#1
15 cm3 of a gaseous hydrocarbon requires 90 cm3 of oxygen for complete
combustion, both volumes being measured at 15 degrees C and 1 atm. The formula of the hydrocarbon is
A C4H6
B C4H8
C C4H10
D impossible to calculate without knowing the molar volume of gases under
these conditions.

I don't really know what to do here. Do I use the ideal gas equation to find the mols? It says the answer is B.
0
3 years ago
#2
pV = nRT
n= (pV)/RT

Try to calculate the moles of each of the elements and see if you get your answer Believe in yourself a bit more, maybe you didn’t need help after all
0
3 years ago
#3
(Original post by SG!YK.VN)
15 cm3 of a gaseous hydrocarbon requires 90 cm3 of oxygen for complete
combustion, both volumes being measured at 15 degrees C and 1 atm. The formula of the hydrocarbon is
A C4H6
B C4H8
C C4H10
D impossible to calculate without knowing the molar volume of gases under
these conditions.

I don't really know what to do here. Do I use the ideal gas equation to find the mols? It says the answer is B.
Well look at the ratio of the volumes. It is 1:6, therefore, I guess it is fine to assume that normal combustion, i.e. full combustion has taken place with CO2 and H2O being formed. In addition, do you get the molar mass of the hydrocarbon, therefore, you could work out the exact number of moles required. This would not be pV=nRT, more so it will be number of moles=mass/Mr
1
3 years ago
#4
Even if it is pV=nRT,
p=101,000 pa
V=1.5*10^-5 m^3
n=?
R=8.314
T=273+15=288 K
n=pV/RT
(101,000*(1.5*10^-5))/(8.314*288)= 6.327..... *10^-4 moles
15cm^3=15g assuming standard density.
Number of moles= 15/Mr
Something does not seem quite right or if my working out is wrong please tell me.
0
3 years ago
#5
(Original post by TurkishMathsHelp)
Well look at the ratio of the volumes. It is 1:6, therefore, I guess it is fine to assume that normal combustion, i.e. full combustion has taken place with CO2 and H2O being formed. In addition, do you get the molar mass of the hydrocarbon, therefore, you could work out the exact number of moles required. This would not be pV=nRT, more so it will be number of moles=mass/Mr
You don’t know the mass though, and mass is different from volume, the easiest equation to use is pV=nRT in this case because you know everything but moles, instead of trying to work out mass from volume
0
3 years ago
#6
(Original post by TurkishMathsHelp)
Even if it is pV=nRT,
p=101,000 pa
V=1.5*10^-5 m^3
n=?
R=8.314
T=273+15=288 K
n=pV/RT
(101,000*(1.5*10^-5))/(8.314*288)= 6.327..... *10^-4 moles
15cm^3=15g assuming standard density.
Number of moles= 15/Mr
Something does not seem quite right or if my working out is wrong please tell me.

Yeah you can’t just assume standard density for one, if you needed it they would tell you in the question what it is. Anyway, you got the moles, then work out the moles of O2, then when you divide moles of O2 by moles of the hydrocarbons you get a ratio of 1:6 of moles of hydrocarbon to moles of O2, so writing out the combustion equation for B, these are the moles used, so the answer is in fact B
1
#7
(Original post by TheMythicalBeast)
Yeah you can’t just assume standard density for one, if you needed it they would tell you in the question what it is. Anyway, you got the moles, then work out the moles of O2, then when you divide moles of O2 by moles of the hydrocarbons you get a ratio of 1:6 of moles of hydrocarbon to moles of O2, so writing out the combustion equation for B, these are the moles used, so the answer is in fact B
Thanks guys I got the answer now 1
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