The conversion of coconut oil to biodiesel occurs by the follow stoichiometric e

Question

The conversion of coconut oil to biodiesel occurs by the follow stoichiometric equation. 1 (coconut oil) + 3 Methanol rightarrow 3 (Biodiesel) + 1 (Glycerol) The conversion of coconut oil is essentially 100% and the products are biodiesel and glycerol. For the small-scale process (assume steady-state), 140L/hr of coconut oil is fed to the reactor, slong with 540 g/hr solid NaOH. Based on previous testing methanol is also fed at a molar flow rate equal to 4.5 times the molar flow rate of coconut oil. A single outlet stream leaves the reactor. The properties of all the components are shown below. What is the mass fraction of the biodiesel in the reactor output? Later stages of the process separate components of this reactor output to provide a stream of essentially pure biodiesel. What is the volumetric flow rate of the stream. What is the concentration of the NaOH catalyst in the output stream?

Explanation / Answer

Since the conversion and the separation is 100%, we can expect 3 moles of biodiesel from one mole of coconut oil, which consume 3 miles of methanol. Therefore there is an excess of 1.5 moles of methanol left in the reactor.

Therefor the mole fraction of biodiesel is 3/(1.5+3+1) = 0.545

moles of NaOH = ( 540/40)= 13.5. It's concentration remains the same during the course of reaction.

Flow rate of biodiesel is same as the flow rate of coconut oil in terms of moles. ( remember it is in steady state)

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