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500 ml of gaseous hydrocarbon when burnt in excess of of O2 gave 2.5 l of CO2 and 3.0 l of H20(vap) the hydrocarbon is
Write the eqn. first and balance it. CxHy + (2x + y/2)O2 =======> xCO2 + (y/2)H2O moles = 0.5/22.4 moles = 2.5/22.4 moles = 3/22.4 Now,from limiting reagent concept, 0.0223/1 = 0.111/xorx = 4.9 approx. 5thus, x = 5and also0.0233/1 = 0.1339*2/y0.0233 = 0.2678/yory = 11.5 that is approx. we woukd take 12.therfore y =12the hydrocarbon is:C5H12.
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