Question

3. As a substitute non-chlorinated solvent for dry cleaning fabrics, bromopropane (C3H7Br) demand has increased. Bromopropane can be synthesized by reacting propanol (C3H8O) with hydrogen bromide (HBr), with water as the second product. A reactor is fed 20 kg/min propanol and 20 kg/min hydrogen bromide. (a) If one of the reactants is completely consumed, what is the composition of the reactor outlet stream? (b) What is the fractional conversion of the excess reactant? (c) What is the extent of reaction? (d) If 60% of the limiting reactant is consumed, what is the yield of bromopropane?

Answer 1

The balanced reaction

C3H7OH + HBr = C3H7Br + H2O

Molar flow of C3H7OH = mass/molecular weight

= (20 kg/min) / (60.09 kg/kmol)

= 0.3328 kmol/min

Molar flow of HBr = mass/molecular weight

= (20 kg/min) / (80.91 kg/kmol)

= 0.2472 kmol/min

Molar flow of HBr < Molar flow of C3H7OH

From the stoichiometry of the reaction

Molar ratio of C3H7OH : HBr = 1 : 1

Limiting reactant = HBr

Excess reactant = C3H7OH

Part a

HBr is completely consumed

At the outlet

Molar flow of HBr = 0

Molar flow of C3H7OH = 0.3328 - 0.2472 = 0.0856 kmol/min

Mass flow of C3H7OH = (0.0856 kmol/min) x (60.09 kg/kmol)

= 5.144 kg/min

Molar flow of C3H7Br = 0.2472 kmol/min

Mass flow of C3H7Br = (0.2472 kmol/min) x (123 kg/kmol)

= 30.406 kg/min

Molar flow of H2O = 0.2472 kmol/min

Mass flow of H2O = (0.2472 kmol/min) x (18 kg/kmol)

= 4.45 kg/min

Total mass = 5.144 + 30.406 + 4.45 = 40 kg/min

Composition in wt%

C3H7OH = 5.144*100/40 = 12.86 %

C3H7Br = 30.406*100/40 = 76.02 %

H2O = 4.45*100/40 = 11.12 %

Part b

Fractional conversion of excess reactant (C3H7OH)

= (mol fed - mol outlet) / (mol fed)

= (0.3328 - 0.0856) / (0.3328)

= 0.7428

Part c

Extent of reaction (e)

C3H7OH balance

Moles out = moles in +[(stoichiometric coefficient) * e]

(0.0856 ) = (0.3328 ) - (1*e)

e = 0.2472 kmol/min

Part d

60% of HBr is consumed

Moles of HBr consumed = 0.2472*0.60 = 0.14832 kmol/min

Moles of C3H7Br produced = 0.14832 kmol/min

Yield of C3H7Br in kg = (0.14832 kmol/min) x (123 kg/kmol)

= 18.243 kg/min