A mixture of methane and ethylene (ethene) in the volume ratio x : y has a total volume of 30ml.On complete combustion it gave 40ml of CO2.If the ratio had been y : x , instead of x : y what volume of CO2 could have been obtained??
Answers
Answered by
7
50 ml co2 will be formed if the ratio is inversed from x:y to y : x...
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Answered by
41
Hey dear,
● Answer -
50 ml
● Explaination -
Combustion of methane is as -
CH4 + 2O2 ---> CO2 + 2H2O
i.e. x CH4 => x CO2
Combustion of ethene is as -
C2H4 + 3O2 ---> 2CO2 + 2H2O
i.e. y C2H4 => 2y CO2
When methane and ethene are burnt together,
x CH4 + y C2H4 --> (x+2y) CO2
Reactants = x + y = 30 ml
Products = x + 2y = 40 ml
Solving this,
Methane = x = 20 ml
Ethene = y = 10 ml
If ratio of methane and ethene is altered, i.e. methane = y = 10 ml and ethene = x = 20 ml
Reactants = y + x = 20 + 10 = 30 ml
Products = y + 2x = 10 + 2×20 = 50 ml
Therefore, later total volume of CO2 produced is 50 ml.
Hope this helps you...
● Answer -
50 ml
● Explaination -
Combustion of methane is as -
CH4 + 2O2 ---> CO2 + 2H2O
i.e. x CH4 => x CO2
Combustion of ethene is as -
C2H4 + 3O2 ---> 2CO2 + 2H2O
i.e. y C2H4 => 2y CO2
When methane and ethene are burnt together,
x CH4 + y C2H4 --> (x+2y) CO2
Reactants = x + y = 30 ml
Products = x + 2y = 40 ml
Solving this,
Methane = x = 20 ml
Ethene = y = 10 ml
If ratio of methane and ethene is altered, i.e. methane = y = 10 ml and ethene = x = 20 ml
Reactants = y + x = 20 + 10 = 30 ml
Products = y + 2x = 10 + 2×20 = 50 ml
Therefore, later total volume of CO2 produced is 50 ml.
Hope this helps you...
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