At 300K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of $O_{2}$ for complete combustion and 40 mL of $C O_{2}$ is formed. The formula of the hydrocarbon is:

C_{4} H_{8}

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C_{4} H_{7} C l

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C_{4} H_{10}

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C_{4} H_{6}

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Solution

The correct option is D $C_{4} H_{6}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$

$1010 (x + \frac{y}{4}) 10 x$

By given data, $10 (x + \frac{y}{4}) = 55$ ...(1)

$10 x = 40$ ...(2)

$∴ x = 4, y = 6 \Rightarrow C_{4} H_{6}$

Hence, option D is correct.

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At 300K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete combustion and 40 mL of CO2 is formed. The formula of the hydrocarbon is:

The correct option is D C4H6CxHy+(x+y4)O2→xCO2+y2H2O1010(x+y4)10xBy given data, 10(x+y4)=55 ...(1)10x=40 ...(2)∴x=4,y=6⇒C4H6Hence, option D is correct.

At 300K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of $O_{2}$ for complete combustion and 40 mL of $C O_{2}$ is formed. The formula of the hydrocarbon is:

The correct option is D $C_{4} H_{6}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$

$1010 (x + \frac{y}{4}) 10 x$

By given data, $10 (x + \frac{y}{4}) = 55$ ...(1)

$10 x = 40$ ...(2)

$∴ x = 4, y = 6 \Rightarrow C_{4} H_{6}$

Hence, option D is correct.