Final answer:
The molar flow rate of unreacted ethane in the product stream is 37 kmol h⁻¹, and the molar flow rate of hydrogen produced is 13 kmol h⁻¹, making option (D) the correct answer.
Step-by-step explanation:
The chemical reaction that describes the pyrolysis of ethane to produce ethylene (C2H4) and hydrogen (H2) is:
C2H6 → C2H4 + H2
Given 50 kmol h-1 of ethane is fed to the reactor, and the product contains 13 kmol of ethylene, we can say that:
- 13 kmol of ethylene is produced from 13 kmol of ethane. Since this is a 1:1 mole ratio in the reaction, 13 kmol of ethane has reacted.
- Thus, the molar flow rate of unreacted ethane in the product stream would be the initial molar flow rate of ethane minus the ethane that has reacted: 50 kmol h-1 - 13 kmol h-1 = 37 kmol h-1 of ethane.
- For each mole of ethane reacted, one mole of hydrogen is produced, which means the molar flow rate of hydrogen in the product stream would be the same as the ethylene, which is 13 kmol h-1.
The correct answer is: (D) Ethane-37 kmol, Hydrogen-13 kmol