$15$ cc of gaseous hydrocarbon required $45$ cc of oxygen for complete combustion and $30$ cc of $C O_{2}$ was formed. The formula of the gaseous compound is:

C_{3} H_{16}

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C_{2} H_{2}

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C_{4} H_{10}

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C_{2} H_{4}

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Solution

The correct option is D $C_{2} H_{4}$
The reaction can be shown as:
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$

Since the ratio of the volumes of oxygen and hydrocarbon used in the combustion is 3, and the ratio of carbon dioxide produced and hydrocarbon used is 2,

$x + \frac{y}{4} = 3$ and $x = 2$

So, we get $y = 4$ .

Therefore, the formula is $C_{2} H_{4}$ .

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15 cc of gaseous hydrocarbon required 45 cc of oxygen for complete combustion and 30 cc of CO2 was formed. The formula of the gaseous compound is:

The correct option is D C2H4The reaction can be shown as: CxHy+(x+y4)O2→xCO2+y2H2OSince the ratio of the volumes of oxygen and hydrocarbon used in the combustion is 3, and the ratio of carbon dioxide produced and hydrocarbon used is 2, x+y4=3 and x=2So, we get y=4.Therefore, the formula is C2H4.

$15$ cc of gaseous hydrocarbon required $45$ cc of oxygen for complete combustion and $30$ cc of $C O_{2}$ was formed. The formula of the gaseous compound is: