Adapt from Question 15 of Chapter 5 of Solen and Harb
A chemical process is used to convert toluene (MW=92) and hydrogen to benzene (MW=78) and methane (CH4) by the reaction
Toluene + H2 → Benzene + CH4
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 H2 (95 mole%) and CH4 (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 H2 and CH4.
If the conversion of toluene in the process is 74.2 %, what is the mass fraction of benzene in the liquid output stream?
Hint: Like many situations in life, this problem may provide more information than you need to answer the given question.
Suggest to follow the below steps
- Calculate the rate of consumption of toluene
- Calculate the left over of toluene in the liquid stream, then convert this amount from mole to mass
- Calculate the rate of benzene formation then convert this amount from mole to mass
- Now you can calculate the mass fraction of benzene in the liquid output stream