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  • 5 L of an alkane requires 25 L of oxygen for its complete combustion.  If all volumes are measured at constant temperature and pressure, the alkane is :Opti

5 L of an alkane requires 25 L of oxygen for its complete combustion.  If all volumes are measured at constant temperature and pressure, the alkane is :

Option: 1

Ethane


Option: 2

Propane


Option: 3

Butane


Option: 4

 Isobutane


Answers (1)

best_answer

Combustion of alkane

\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}+\left(\frac{3 \mathrm{n}+1}{2}\right) \longrightarrow \mathrm{nCO}_{2}+(\mathrm{n}+1) \mathrm{H}_{2} \mathrm{O}

From the above reaction,

No. of moles of alkane =1

\text { No. of moles of oxygen }=\frac{3 n+1}{2}

From ideal gas equation,

PV=nRT

At same temperature and pressure,

\begin{array}{l} \mathrm{V} \propto \mathrm{n} \\ \Rightarrow \mathrm{V}_{(\mathrm{alkane})} \mathrm{n}_{(\mathrm{oxygen})}=\mathrm{V}_{(\operatorname{oxygen})} \mathrm{n}_{(\mathrm{alkane})} \cdots \cdots(1) \end{array}

Given:-

Volume of alkane =5L

Volume of oxygen =25L

From equation (1), we have

\begin{array}{l} 5 \times\left(\frac{3 n+1}{2}\right)=25 \times 1 \\ \frac{3 n+1}{2}=\frac{25}{5} \\ \Rightarrow 3 n+1=5 \times 2 \\ \Rightarrow 3 n=10-1 \\ \Rightarrow n=3 \end{array}

\text { Thus the alkane is } \mathrm{C}_{3} \mathrm{H}_{2 \times 3+2}=\mathrm{C}_{3} \mathrm{H}_{8} \text { , it is propane. }

Posted by

sudhir.kumar

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