JEE MAIN - Chemistry (2019 - 10th April Morning Slot - No. 12)
At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete
combustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is :
C4H7Cl
C4H6
C4H8
C4H10
Explanation
CxHy + $$\left( {x + {y \over 4}} \right)$$O2 $$ \to $$ xCO2 + $${y \over 2}$$H2O
According to the above equation,
1 mL of hydrocarbon = x mL of CO2 is produced
$$ \therefore $$ 10 mL of hydrocarbon = 10x mL of CO2 is produced.
According to question,
From 10 mL of hydrocarbon 40 mL of CO2 is produced.
$$ \therefore $$ 10x = 40
$$ \Rightarrow $$ x = 4
Also from the above equation,
1 mL of CxHy react with $$\left( {x + {y \over 4}} \right)$$ mL of O2.
$$ \therefore $$ 10 mL of CxHy react with 10$$\left( {x + {y \over 4}} \right)$$ mL of O2.
According to question,
10$$\left( {x + {y \over 4}} \right)$$ = 55
$$ \Rightarrow $$ $$40 + {{10y} \over 4} = 55$$
$$ \Rightarrow $$ y = 6
$$ \therefore $$ Hydrocarbon is C4H6
According to the above equation,
1 mL of hydrocarbon = x mL of CO2 is produced
$$ \therefore $$ 10 mL of hydrocarbon = 10x mL of CO2 is produced.
According to question,
From 10 mL of hydrocarbon 40 mL of CO2 is produced.
$$ \therefore $$ 10x = 40
$$ \Rightarrow $$ x = 4
Also from the above equation,
1 mL of CxHy react with $$\left( {x + {y \over 4}} \right)$$ mL of O2.
$$ \therefore $$ 10 mL of CxHy react with 10$$\left( {x + {y \over 4}} \right)$$ mL of O2.
According to question,
10$$\left( {x + {y \over 4}} \right)$$ = 55
$$ \Rightarrow $$ $$40 + {{10y} \over 4} = 55$$
$$ \Rightarrow $$ y = 6
$$ \therefore $$ Hydrocarbon is C4H6
Comments (0)
