A ternary mixture of n-buatane, 1-butene and furfural is analyzed to find the co

Question

A ternary mixture of n-buatane, 1-butene and furfural is analyzed to find the content of each in it. The mixture is stripped off with the help of carbon dioxide without appreciable entrainment of furfural due to its very low vapor pressure. The stripped gases are passed through an absorber column in which CO2 is absorbed in 25% (by mass) KOH solution. The mixture of hydrocarbons, saturated with water vapor is collected in a measuring burette. Data: Sample mass 6.5g volume of saturated gases collected at 296K and 102.5kPa=415.1 ml n-Butane present in the hydrocarbon in the bureete-43,.1mbly Find the analysis of the mixture (both on mole basis and mass basis) Data: Vapour pressure of water over 25% KOH solution at 296.8K = 2.1 75kPa (Stchiometry by B.i.Bhatt and Vora)

Explanation / Answer

Sample mass (n-butane + 1-butene + furfural) = 6.5g

Vapor pressure of water over 25% KOH solution = 2.1 75kPa

So, partial pressure of n-butane + 1-butene = 102.5kPa - 2.1 75kPa = 100.325 kPa

Now calculate the moles of (n-butane + 1-butene) by using ideal gas equation

PV = n RT

Where P = 100.325 kPa

V = 415.1 ml = 415.1/1000 = 0.4151 L

n = moles of (n-butane + 1-butene)

R = Gas constant =8.314 L kPa K1 mol1

T = 296K

n = PV/ RT

= (100.325 x 0.4151) /(8.314 x 296) = 0.01692 moles

Calculate the n-butane in mixture of hydrocarbon

n-Butane in the hydrocarbon mixture = 43.1 mol %

Fraction of n-Butane in the hydrocarbon mixture = 0.431

Moles of n-Butane in the hydrocarbon mixture = 0.01692 x 0.431 = 0.00729 moles

Molar mass of n-Butane (C4H10) = 58 g/mol

Mass of n-Butane in the hydrocarbon mixture = 0.00729 moles x 58 g/mol = 0.423 g

Moles of 1-Butene in the hydrocarbon mixture = 0.01692 moles - 0.00729 moles = 0.00963 moles

Molar mass of 1-Butene (C4H8) = 56 g/mol

Mass of 1-Butene in the hydrocarbon mixture = 0.00963 moles x 56 g/mol = 0.539 g

Mass of furfural = mass of (n-butane + 1-butene + furfural – mass of(n-butane + 1-butene)

= 6.5 g – (0.423 g + 0.539 g) = 5.538 g

Molar mass of furfural (C5H4O2) = 96 g/mol

Moles of furfural in the liquid mixture = 5.538 g /96 g/mol = 0.05768 moles

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