At 300 K and 1 atmsopheric pressure, 10 mL of a hydrocarbon required 55 mL of O_{2}

for complete combustion, and 40 mL of CO_{2}  is formed. The formula of the hydrocarbon is:

Answers (1)

The eqn of hydrocarbon can be written as : 

C_xH_y+(x+\frac{y}{4})O_2\rightarrow xCO_2+\frac{y}{2}H_2O

Since , 10mL of C_xH_y produces 40mL of CO_2 & 

             1mL of C_xH_y produces x mL of CO_2 

                     \therefore x=\frac{40}{10}=4

Now,

         10mL of C_xH_y  requires 55mL of O_2

       \therefore 1mL of C_xH_y  requires (\frac{55}{10})mL of O_2

\therefore x+\frac{y}{4}=\frac{55}{10}

=> \frac{y}{4}=\frac{55}{10}-x

=> y=4(5.5-x)

=>y=4(5.5-4)

=>y=6

\therefore Hydrocarbon=C_4H_6

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