10.0^cm3 of propane was sparked with 60 of oxygen at s.t.p.what is the volume of residu gas after the reaction is cooled
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I need the solution to that question
You have no units for the O2. I assume that is 60 cc.
C3H8 + 5O2 ==> 3CO2 + 4H2O
10 cc C3H8 will need 50 cc O2 to react completely and will produce 30 cc CO2 gas and 40 cc H2O gas (vapor at the combustion temperature). There will be 10 cc O2 remaining un-reacted. After cooling the 40 cc H2O vapor will not be a gas at stp. So the gas volume will be
0 cc C3H8 + 10 cc O2 un-reacted + 30 cc CO2 produced.
I'm not exactly sure what you mean by residual gas. If you mean the volume of the gas at the end of the reaction that will be 10 + 30 = 40 cc. If you mean the volume of the un-reacted O2 gas it will be 10 cc.
C3H8 + 5O2 ==> 3CO2 + 4H2O
10 cc C3H8 will need 50 cc O2 to react completely and will produce 30 cc CO2 gas and 40 cc H2O gas (vapor at the combustion temperature). There will be 10 cc O2 remaining un-reacted. After cooling the 40 cc H2O vapor will not be a gas at stp. So the gas volume will be
0 cc C3H8 + 10 cc O2 un-reacted + 30 cc CO2 produced.
I'm not exactly sure what you mean by residual gas. If you mean the volume of the gas at the end of the reaction that will be 10 + 30 = 40 cc. If you mean the volume of the un-reacted O2 gas it will be 10 cc.
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