At 300 K and 1 text{atm} pressure,10 mL of | vedclass

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(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by: $C_xH_y + (x + \frac{y}{4})O_2 	o xCO_2 + \frac{y}{2}H_2O$ According to Avogadro'

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$C_4H_6$

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(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by: $C_xH_y + (x + \frac{y}{4})O_2 	o xCO_2 + \frac{y}{2}H_2O$ According to Avogadro's Law,at constant temperature and pressure,the volume of gases is proportional to their moles. Given: Volume of $C_xH_y = 10 \ mL$,Volume of $O_2 = 55 \ mL$,Volume of $CO_2 = 40 \ mL$. From the stoichiometry: $10x = 40 \implies x = 4$ $10(x + \frac{y}{4}) = 55$ $x + \frac{y}{4} = 5.5$ Substituting $x = 4$: $4 + \frac{y}{4} = 5.5$ $\frac{y}{4} = 1.5$ $y = 6$ Thus,the formula of the hydrocarbon is $C_4H_6$.

At $300 \ K$ and $1 \ \text{atm}$ 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

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