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

Butane

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Isobutane

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Ethane

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Propane

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Solution

The correct option is D Propane
$C_{n} H_{2 n + 2} + (\frac{3 n + 1}{2}) O_{2} \to n C O_{2} + (n + 1) H_{2} O$
$5 L$ $25 L$

Since volume are measured at constant $T$ and $P$ .

Therefore,volume is directly proportional mole.

Since,

$n_{a l k a n e} = (\frac{2}{3 n + 1}) \times n_{o_{2}}$
Implies that

$5 = \frac{2}{3 n + 1} \times 25$

$n = 3.$
Hence the alkane is propane $(C_{3} H_{8})$ .

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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, then the alkane is:

The correct option is D PropaneCnH2n+2+(3n+12)O2→nCO2+(n+1)H2O 5L 25LSince volume are measured at constant T and P.Therefore,volume is directly proportional mole.Since,nalkane=(23n+1)×no2Implies that5=23n+1×25 n=3.Hence the alkane is propane (C3H8).

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