A chemical process is used to convert toluene (MW=92) and hydrogen to benzene (M

Question

A chemical process is used to convert toluene (MW=92) and hydrogen to benzene (MW=78) and methane (CH_4) by the reaction Toluene + H_2 rightarrow Benzene + CH_4 Two streams enter the process. The first input stream is pure liquid toluene, which enters at a rate of 40 mol/s. The second input stream is a mixture of H_2 (95 mole%) and CH_4 (5 mole The flow rate of H2 in the second stream is equal to 200 mol/s. Two streams leave the process. The first output stream contains only liquid benzene (product) and toluene (unconverted reactant). The second output stream contains gaseous H_2 and CH_4. If the conversion of toluene in the process is 80.8 %. what is the mass fraction of benzene in the liquid output stream?

Explanation / Answer

First stream isToluene and its low rate = 40 mol/s

There is 95 mole % H2 in the second stream whose flow rate is 200 mol/s

Second stream is 200/0.95=210.5 mole/s

Toluene is the limiting reactant

Conversion of toluene is 80.8%, moles of toluene converted = 0.808*40 =32.32 mol/s

Benzene formed = 32.32 mol/s   tolune unconverted = 40-32.32=7.68 mol/s

Output stream contains liqudi stream containing liqujid Benzene whose flow rate is =32.32 mol/s and toluene which is unconverted = 7.68 mol/s

Mass = moles* molecular weight

Mass of Benzene = 32.32*78=2521 gm/s and mass of toluene = 7.68*92 =707 gm/s

Total mas of liquds = 707+2521=3228 gm/s

Mass fraction of Benzene in the liquid stream = mass of Benzene /total mass= 2521/3228=0.79

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