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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:
Option 1)
$C_{4}H_{10}$
Option 2)
$C_{4}H_{6}$
Option 3)
$C_{4}H_{7}Cl$
Option 4)
$C_{4}H_{8}$

Answers (1)

best_answer

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 $O_2$

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

=> $\frac{y}{4}=\frac{55}{10}-x$

=> $y=4(5.5-x)$

=> $y=4(5.5-4)$

=> $y=6$

$\therefore Hydrocarbon=C_4H_6$

So, option (2) is correct.

Option 1)

$C_{4}H_{10}$

Option 2)
$C_{4}H_{6}$
Option 3)

$C_{4}H_{7}Cl$

Option 4)

$C_{4}H_{8}$

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