$V mL$ of a hydrocarbon $C_{x} H_{y}$ requires $6 V mL$ of oxygen for complete combustion & forms $4 V mL$ of $C O_{2} .$ The value of $y$ is .

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Solution

$\begin{matrix} C_{x} H_{y} & + (x + \frac{y}{4}) O_{2} \to & x C O_{2} + & (\frac{y}{2}) H_{2} O (l) Volume-Volume Analysis & V & 6 V & 4 V \end{matrix}$

$\frac{V_{C_{x} H_{y}}}{1} = \frac{V_{C O_{2}}}{x}$
$\frac{v}{1} = \frac{4 v}{x} x = 4$

$\frac{V_{C_{x} H_{y}}}{1} = \frac{V_{O_{2}}}{x + \frac{y}{4}}$

$\frac{v}{1} - \frac{6 v}{x + \frac{y}{4}}$

$x + \frac{y}{4} = 6$
$4 + \frac{y}{4} = 6$
$\frac{y}{4} = 2$
$y = 8$
$Formula will be C_{4} H_{8}$ .

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V mL of a hydrocarbon CxHy requires 6V mL of oxygen for complete combustion & forms 4V mL of CO2. The value of y is .

CxHy+(x+y4)O2→xCO2+(y2)H2O(l)Volume-Volume AnalysisV6V4V VCxHy1=VCO2x v1=4vxx=4 VCxHy1=VO2x+y4 v1−6vx+y4 x+y4=6 4+y4=6 y4=2 y=8 Formula will be C4H8.

$V mL$ of a hydrocarbon $C_{x} H_{y}$ requires $6 V mL$ of oxygen for complete combustion & forms $4 V mL$ of $C O_{2} .$ The value of $y$ is .