Write some of the hydrocarbon rxns with oxygen. All give CO2 + H2O. The problem would be easier if you were given cc H2O (as a gas) but if you don't have that I would do this.
CH4 + 2O2 ==> C2 + 2H2O
C2H6 + 7O2 ==> 4CO2 + 6H2O
C3H8 + 5O2 --> 3CO2 + 4H2O
2C4H10 + 13O2 ==> 8CO2 + 10H2O
etc. Now what did you have?
CxHy + O2 ==> xCO2 + (y/2)H2O
20 cc CxHy
O2 used must be 90-20 = 70 cc.
CO2 produced = 40 cc
The ratio of these numbers is
CxHy + O2 ....> xCO2 + (y/2)H2O
..20...70.......40........? OR
1.....3.5.......2.......?
That means x must 2 but what is H2O. If you have it in the problem that makes it easier. If you don't have it then you show that of all of the combustions I wrote at the beginning the one with C2H6 is the only one that fits. That is the ONLY one with CO2 twice the hydrocarbon and reactions I didn't write for saturated hydrocarbons become more divergent. Note that H2O in C2H6 (when combusted) is 3x that of the hydrocarbon. So y/2 = 3 and y = 6 and CxHy is C2H6.
20cm(cube) of a gaseous hydrocarbon were mixed with 90cm(cube) oxygen, and the mixture exploded. At room temperature, 60cm(cube) of gas were left. 40cm(cube) of the gas(cabon(4)oxide) were absorbed by sodium hydroxide, leaving 20cm(cube) of oxygen. find the empirical formular of the hydrocarbon.
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