15 ml of a gaseous hydrocarbon required 75 ml of O2 for complete combustion and it produced 45 ml of CO2 (All volumes are measured under identical conditions).The hydrocarbon is

a)C5H12 b)C3H6 c)C4H10 d)C3H8

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Solution

C_xH_y + (2x+y/2)/2 O₂ → xCO₂ + y/2 H₂O Find the volume of O₂ used = 75ml.. Volume of C_xH_y = 15ml that means 4x + y = 20 so possible values of x and y x y 3 8 15 ml of C_xH_y will produce 45 ml of CO₂ volume of product = volume of CO₂ + volume of left over air 327 = volume of CO₂ + 357 - 75 volume of CO₂ = 45ml that means x = 3..so the case we took is ryt..hence the answer is C₃H₈

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and

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