At 300 K and 1 atm , 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20 % O 2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :

Question

At 300 K and 1 atm , 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20 % O 2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is : (A) C3H6 (B) C3H8 (C) C4H8 (D) C4H10

Tardigrade Admin · Accepted Answer

underset15 textmLCX HY + underset15(x + (y/4)) textmL(x+(Y/4))O2arrow underset15 text x mLxCO2+(Y/2)H2O VO2 = (20/100) × 375 = 75 mL = 15 (x + (y/4)) ⇒ x + (y/4) = 5 ⇒ C3H8

$C_{3} H_{6}$

$C_{3} H_{8}$

$C_{4} H_{8}$

$C_{4} H_{10}$

$15 mL C_{X} H_{Y} + 15 (x + \frac{y}{4}) mL (x + \frac{Y}{4}) O_{2} \to 15 x mL x C O_{2} + \frac{Y}{2} H_{2} O$

$V_{O_{2}} = \frac{20}{100} \times 375 = 75 m L = 15 (x + \frac{y}{4})$

$\Rightarrow x + \frac{y}{4} = 5$

$\Rightarrow C_{3} H_{8}$

C3​H6​

C3​H8​

C4​H8​

C4​H10​

15mLCX​HY​​+15(x+4y​)mL(x+4Y​)O2​​→15 x mLxCO2​​+2Y​H2​O VO2​​=10020​×375=75mL=15(x+4y​) ⇒x+4y​=5 ⇒C3​H8​

$C_{3} H_{6}$

$C_{3} H_{8}$

$C_{4} H_{8}$

$C_{4} H_{10}$

$15 mL C_{X} H_{Y} + 15 (x + \frac{y}{4}) mL (x + \frac{Y}{4}) O_{2} \to 15 x mL x C O_{2} + \frac{Y}{2} H_{2} O$

$V_{O_{2}} = \frac{20}{100} \times 375 = 75 m L = 15 (x + \frac{y}{4})$

$\Rightarrow x + \frac{y}{4} = 5$

$\Rightarrow C_{3} H_{8}$