
At 300K and 1 atm, 15 ml of a gaseous hydrocarbon requires 375 ml air containing 20% 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 :
A.
B.
C.
D.
Answer
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Hint: Hydrocarbon is defined as an organic compound that contains entirely hydrogen and carbon atoms. They are classified into four categories such as alkane, alkenes, alkynes, and aromatic .
Complete step by step answer:
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 requires can be calculated as:
100 ml of air contain=20 ml of oxygen.
1 ml of air contain= of oxygen
375 ml of air contain=
Now find out the ratio of the volume of oxygen used for combustion to the volume of hydrocarbon as:
This ratio can be obtained in the case of propane as:
The chemical reaction for the combustion of propane as:
Here water obtained is in liquid form. The volume of hydrocarbon is given as 15 ml. Hence we can say 15 ml of propane on combustion gives = of carbon dioxide gas.
And the volume of oxygen used is 75 ml. Thus the volume of remaining air except oxygen is . Hence the total volume will be . Here some quantity of carbon dioxide can dissolve in water hence the total volume occupied by the gas is less than . Hence we can say the total volume occupied by gas is .
Hence the correct answer is option A.
Note:
Propane is commonly used as a fuel. It is one of the groups of liquefied petroleum gas. It is commonly used for space and water heating. Since it is used as fuel in vehicles hence it is known as propane autogas.
Complete step by step answer:
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 requires can be calculated as:
100 ml of air contain=20 ml of oxygen.
1 ml of air contain=
375 ml of air contain=
Now find out the ratio of the volume of oxygen used for combustion to the volume of hydrocarbon as:
This ratio can be obtained in the case of propane as:
The chemical reaction for the combustion of propane as:
Here water obtained is in liquid form. The volume of hydrocarbon is given as 15 ml. Hence we can say 15 ml of propane on combustion gives =
And the volume of oxygen used is 75 ml. Thus the volume of remaining air except oxygen is
Hence the correct answer is option A.
Note:
Propane is commonly used as a fuel. It is one of the groups of liquefied petroleum gas. It is commonly used for space and water heating. Since it is used as fuel in vehicles hence it is known as propane autogas.
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