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Q. If a $LPG$ cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of $1\, kg$ of it will be

UPSEEUPSEE 2013Thermodynamics

Solution:

The standard enthalpy of combustion of butane $C _{4} H _{10}$ (both $n$ -and iso butanes are main constituents of cooking gas). representing the combustion of 1 mole of butane in the presence of oxygen, may be representing represented as follows

$\underset{\text{1 mol}}{{C _{4} H _{10}( g )}}+\underset{\frac{13}{12}mol}{\frac{13}{2} O _{2}(g) }\rightarrow 4 CO _{2}(g)+5 H _{2} O (l)$;

$\Delta_{c} H^{\circ}=-2658.0 \,kJ \,mol ^{-1}$

$(58 \,g) \left(\frac{13}{2} \times 32=208 g \right)$

$\therefore \, O _{2}$ required for $1\, kg$ butane

$=\frac{208}{58} \times 1000 \,g $

$=3.58 \times 10^{3} \,g$