Question

Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H34O2)/mol, and 0.59 mol linoleic acid (C18H32O2)/mol. To create margarine from sunflower oil, the liquid oleic and linoleic acids are hydrogenated in the presence of a metal catalyst to form solid stearic acid. At a margarine production facility, a stream of 335.0 mol sunflower oil/hr is fed into a reactor. The fresh hydrogen source available is a mixture containing 0.95 mol H2/mol and 0.050 mol N2/mol. (N2 is inert in this process.) To insure complete hydrogenation, the hydrogen gas mixture is introduced into the reactor in 65.0% excess. All reaction products leave the reactor and proceed to a separator where palmitic and stearic acids are separated from the hydrogen gas mixture. The hydrogen gas mixture is recycled to the fresh feed stream, and 140.0 mol gas/hr of this recycle stream is purged to prevent the buildup of nitrogen gas. What is the composition of the purge stream?

Answer 1

Answer:

$x_H=0.882$

$x_N=0.118$

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: $C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)$

Linoleic: $C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)$

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

Base of caculation: 1 mol of sunflower oil

For oleic acid: $n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}$

$n_{Holeic}=frac{90.45 mol H_2}{hr}$

For linoleic acid: $n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}$

$n_{Holeic}=frac{395.3 mol H_2}{hr}$

$n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}$

$n_{Htotal}=frac{485.75 mol H_2}{hr}$

Applying the excess:

$n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol$

Nitrogen: $n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}$

$n_N= 42.2 mol N$

After the reactions:

$n_H=801.48 mol-485.75mol=315.73 mol$

and the nitrogen is inert.

Purge stream:

$n_total=42.2+315.73 mol=357.93 mol$

$x_H=\frac{315.73mol}{357.93mol}=0.882$

$x_N=\frac{42.2mol}{357.93mol}=0.118$