At 300 K and 1 atm, 15 mL of a gaseous hyd | vedclass

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(D) The combustion reaction is: $C_xH_y + (x + \frac{y}{4}) O_2 ightarrow x CO_2 + \frac{y}{2} H_2O_{(l)}$Volume of $O_2$ used $= 375 \ mL 	imes 0.

Vedclass · Accepted Answer

$C_3H_8$

Vedclass · Answer

(D) The combustion reaction is: $C_xH_y + (x + \frac{y}{4}) O_2 ightarrow x CO_2 + \frac{y}{2} H_2O_{(l)}$Volume of $O_2$ used $= 375 \ mL 	imes 0.20 = 75 \ mL$.Volume of $N_2$ in air $= 375 \ mL - 75 \ mL = 300 \ mL$.From the stoichiometry of combustion: $15(x + \frac{y}{4}) = 75 \implies x + \frac{y}{4} = 5 \implies 4x + y = 20$.The final volume of $345 \ mL$ consists of $CO_2$ produced and the unreacted $N_2$.Volume of $CO_2 = 345 \ mL - 300 \ mL = 45 \ mL$.Since $15 \ mL$ of hydrocarbon produces $45 \ mL$ of $CO_2$, $15x = 45 \implies x = 3$.Substituting $x = 3$ into $4x + y = 20$: $4(3) + y = 20 \implies 12 + y = 20 \implies y = 8$.Thus, the hydrocarbon is $C_3H_8$.

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