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A gas mixture of 6 L of propaneand butane on complete combustion at 25 Celsius produce 20 L co2 Volcker of butane present initially in mixture is
Since process has took as complete combustion concepts of POAC can be applied here.No. of total carbons in Reactant = > 7and let V be the volume of butane present in the initial mixture.Therefore, Volume of propane = > 6 – VHence,V * 4 + (6- V)*3 = 1 * 20Solving for V we get,4V – 3V = 20 – 18V = 2 litre.Hence butane is 2 litre in capacity.
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