The following reaction occurs in a plug flow reactor: C_4H_20_3 + 2C0_2+ 2H_20 r

Question

The following reaction occurs in a plug flow reactor: C_4H_20_3 + 2C0_2+ 2H_20 rightarrow C_6H_6 + 9/20_2 The first reactant is known as maleic anhydride (MA) and is used in the production of benzene (B). A feed stream contains 1.250 kmol MA/h. 10,000 kg CO_2/h. and 3.000 kmol H_20/h. Assume only gas phase present and that the volume is variable. What is the limiting reactant? Provide a mole balance equation and rate of generation for the limiting reactant only. Complete a stoichiometric table for this process.

Explanation / Answer

a.

C4H2O3 + 2CO2 + 2H2O ---- > C6H6 + 9/2 O2

First calcualte the number of moles of all recatants

MA/h = 1250 kmol = 1250,000 mol

10,000 kg CO2 /h= 10,000 ,000 g CO2 /h

Molar mass of CO2= 44 g/ mol

CO2/h mol = 10,000 ,000 g CO2 /h44 g/ mol

= 227,272.7 mol

3,000 kmol H2O /h = 3,000,000 mol H2O /h

Limiting agent is completely used in the reaction and dermines the amount of the product.

Here CO2 is limting agent.

227,272.7 mol CO2 *1 mole C6H6 / 2 mol CO2=

= 11363.35 mole C6H6

b

C4H2O3 + 2CO2 + 2H2O ---- > C6H6 + 9/2 O2

The rate of generating for the limiting agent CO2 is 11363.35 mole C6H6/h.

c.

C4H2O3 + 2CO2 + 2H2O ---- > C6H6 + 9/2 O2

227,272.7 mol CO2 *1 mole C6H6 / 2 mol CO2=

= 11363.35 mole C6H6

1250,000 mol MA *1 mole C6H6 / 1 mol MA=

= 1250,000 mol C6H6

3,000,000 mol H2O *1 mole C6H6 / 2 mol H20=

= 1500,000 mole C6H6

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