Question
Asked Sep 3, 2019
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34.4 gallon of toluene, 85.4 lbm of ethyl acetate, and 3.26 ft cubic of ethanol is poured into a tank. Find the mass fraction of each component.

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Expert Answer

Step 1

Given information:

Volume of Toluene (vt)= 34.4 gallon

Mass of Ethyl acetate (me) = 85.4 lbm

Volume of Ethanol (vEtOH) = 3.26 ft3

The mass fraction for all above components, poured into a tank, is to be calculated. Since, the volume is given for Toluene and Ethanol, the first step is to calculate the mass (in kg) using the density of these components.

Density of Toluene ρt = 867 kg/m3

Density of Ethanol ρEtOH = 789 kg/m3

As the density is in SI units, the following unit conversions are to be considered while calculating mass of all the components.

Unit conversion:

1 gallon = 0.0038 m3

1 ft3 = 0.028 m3

1 lbm = 0.4536 kg

The unit conversion of mass of Ethyl acetate to kg is given in equation (1). The unit conversion of volume of Toluene and Ethanol to m3 is shown in equation (2) and (3) respectively.

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me 85.4 lbmx 0.4536 kg 1lbm (1) 38.72 kg 34.4 gallonx 0.0038 m* 1 gallon (2) =0.13 m3 E1OH 3.26 ft3x0.028 m3 1 f3 (3) -0.091 m2

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Step 2

It is known that density of a component is the ratio of its mass to volume. So, the mass for Toluene and Ethanol is calculated in equation (4) and (5) respectively.

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Mass of Toluene (m, PtVolume of Toluene (v,) 867 kg/m3 0.13 m m4113.33 kg Mass of Ethan ol (m E10H' = 1", (4) PE1OH Volume of Ethanol (vEtOH 789 kg/m3= EtOH 0.091 m3 (5) m "EtOH =72.02 kg

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Step 3

Mass fraction for an ith component in a given mixture of n components is given by equation (6). The total mass of the...

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Mass of component i Total mass of the mixture Total mass of mixture= mOH +"e+mt (6) 72.02 kg+38.72 kg+113.33 kg = 224.07 kg (7)

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