1
Question
At 300K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion, the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure. The formula of the hydrocarbon is :
At 300K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion, the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure. The formula of the hydrocarbon is :
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Solution
The correct option is A C3H8
Here given that for the complete combustion of 15 mL of a gaseous hydrocarbon 375 mL of air containing
20% of oxygen by volume is required. Thus the volume of oxygen required can be calculated as:
100 mL of air contains 20 mL of oxygen.
So,
1 mL of air contains 20100 mL of oxygen
375 mL of air contain =20100×375=75 mL
The ratio of the volume of oxygen used for combustion to the volume of hydrocarbon is:
volume of oxygen usedvolume of hydrogen=7515=5:1
This ratio can be obtained in the case of propane as:
The chemical reaction for the combustion of propane
as:
C3H8+5O2→3CO2+4H2O
Here water obtained is in liquid form. The volume of hydrocarbon is given as 15 mL. Hence, 15 mL of
propane on combustion gives =3×15=45mL of carbon dioxide gas
And the volume of oxygen used is 75 mL. Thus the volume of remaining air except oxygen is (375−75)=300 mL.
Hence the total volume will be 300+45=345mL. Some quantity of carbon dioxide can dissolve in water hence the total volume occupied by the gas is less than 345 mL.
So, the total volume occupied by gas is 330 mL
Hence, the correct answer is option A.
Here given that for the complete combustion of 15 mL of a gaseous hydrocarbon 375 mL of air containing
20% of oxygen by volume is required. Thus the volume of oxygen required can be calculated as:
100 mL of air contains 20 mL of oxygen.
So,
1 mL of air contains 20100 mL of oxygen
375 mL of air contain =20100×375=75 mL
The ratio of the volume of oxygen used for combustion to the volume of hydrocarbon is:
volume of oxygen usedvolume of hydrogen=7515=5:1
This ratio can be obtained in the case of propane as:
The chemical reaction for the combustion of propane
as:
C3H8+5O2→3CO2+4H2O
Here water obtained is in liquid form. The volume of hydrocarbon is given as 15 mL. Hence, 15 mL of
propane on combustion gives =3×15=45mL of carbon dioxide gas
And the volume of oxygen used is 75 mL. Thus the volume of remaining air except oxygen is (375−75)=300 mL.
Hence the total volume will be 300+45=345mL. Some quantity of carbon dioxide can dissolve in water hence the total volume occupied by the gas is less than 345 mL.
So, the total volume occupied by gas is 330 mL
Hence, the correct answer is option A.
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