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CxHy + (2x+y/2)/2 O2 → xCO2 + y/2 H2OFind the volume of O2 used = 75ml..Volume of CxHy = 15mlthat means 4x + y = 20so possible values of x and yx y3 815 ml of CxHy will produce 45 ml of CO2volume of product = volume of CO2 + volume of left over air327 = volume of CO2 + 357 - 75volume of CO2 = 45mlthat means x = 3..so the case we took is ryt..hence the answer is C3H8
PS : The trick in the problem is that..water will not be in gaseous form at NTP
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