Question

Enter your answer in the provided box. Acrylic acid (CH2═CHCOOH) is used to prepare polymers, adhesives, and paints. The first step to make acrylic acid involves the vapor-phase oxidation of propylene (CH2═CHCH3) to acrolein (CH2═CHCHO). This step is carried out at 330°C and 2.50 atm in a large bundle of tubes around which circulates a heat-transfer agent. The reactants spend an average of 1.80 s in the tubes, which have a void space of 92.1 ft3. How many pounds of propylene must be added per hour in a mixture whose mole fractions are 0.0700 propylene, 0.3500 steam, and 0.5800 air?

Answer 1

Answer: m = 1710.35 pounds per hour

Explanation: Since the step involves a gaseous state, it can be used the ideal gas formula, given by:

PV=nRT

For this question, R will be 8.31451 m³Pa/K.mol, which means the variables need a change of unit:

T = 330°C + 273 = 603K

P = 2.5atm*101325 = 253312.5Pa

V = 92.1ft³*0.0283 = 2.60643m³

Calculating mols:

PV=nRT

$n=\frac{253312.5*2.60643}{8.31451*603}$

n = 131.7 mols

The reactant (propylene) spends 1.80 seconds in the tubes and 1 hour is 3600 seconds, so:

$\frac{131.7*3600}{1.8}$ = 263377.5 mol/h

In the mixture, the proportion of propylene is 0.07, then:

263377.5*0.07 = 18436.4 mol of propylene in the mixture

Mol is related to mass and molar mass by the following formula:

$n=\frac{m}{M}$

m = nM

Molar Mass of propylene is 42.08g/mol:

m = 18436.4*42.08 = 775803.712g

Changing into pounds:

$m=\frac{775803.712}{453.6}$

m = 1710.35lbs

It must be added 1710.35lbs per hour in the mixture.