The Fischer-Tropsch process is a reaction in mixture of and hydrogen react to fo

Question

The Fischer-Tropsch process is a reaction in mixture of and hydrogen react to form The length of the hydrocarbon chain is dependent on pressure and temperature, but the reaction is not very selective and often product form. Carbon and hydrogen gas are fed into a reactor at 200.0°C and 1 atm. The product gas exits at 200.0 C, 1 atm and contains 16.7 mo/hr of n-butane (C4H10), 18.3 mollhr of n-pentane (C5H12), water vapor, and unreacted carbon monoxide and hydrogen gas. If 78.0% of the incoming carbon monoxide gas and 73.0% of the incoming hydrogen gas react, find the heat flow of gas into or out of the reactor by completing the following First label process below. Label any if are not given or implied above. leave any blank spaces. names are given for reference n1 unknown mol CO/hr 2000 Reactor °C 200.0 °C nz unknown mol H2/hr 0 0 0 ng unknown mol CO/hr 200.0 °C n4 unknown mol H2/hr ns- unknown mol H20 (g)hr ns 16,7 mol C4H10 (gyhr kJ/hr nr 18.3 mol C5H12 (gyhr 73.0 78.0 18.3 16.7 200.00) unknown Next, perform mass balances around the reactor to find the unknown fiow rates. What is n, the flow rate of carbon monoxide into the reactor?

Explanation / Answer

The reactions are

4CO +9H2 -----àC4H10+ 4H2O (1) and 5CO + 11H2 -----à C5H12+ 5H2O (2)

For reaction (1), Moles of C4H10 formed= 16.7, mole of CO required = 4*16.7 =66.8, moles of H2 required= 9*16.7= 150.3 moles.

For reaction (2), moles of CO required = 5*18.3= 91.5, moles of H2 required= 11* 18.3= 201.3

Total moles of CO= 66.8+91.5= 158.3, but CO is 78% converted,

Co entering the reactor = 158.3/0.78= 203 moles/hr

H2 supplied = 150.3+201.3= 351.6, H2 supplied got only 73% converted, H2 entering the reactor= 351.6/0.73=482 moles/hr

CO leaving the reactor = 203*0.22= 44.66 moles/hr, H2 leaving the reactor = 482*0.27= 130 moles/hr

Moles of water prodiced = 4*16.7 +5*18.3= 158.3 moles/hr

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