A reactor is charged with 1,000 [ton] of A (solid) and is continuously fed with

Question

A reactor is charged with 1,000 [ton] of A (solid) and is continuously fed with a liquid stream of 1 [ton/h] of B with 10 % w/w of contaminant D.2A + 5B rightarrow 40C A + D rightarrow 24E + 2C (both reactions have 100 % conversion) The products of the reactions are gases, and it is desirable to remove 90 % of the contaminating E from this stream, so an adsorption tower will be installed to remove contaminant E. The tower can withdraw 0.05 [kg] E/kg of adsorbent. What is the molecular weight of E Calculate the load of adsorbent (kg) required if the adsorbent must be regenerated every three reactor loads. MW_A = 150; MW_B = 60; MWC = 15; MW_D = 24 MW_E = 6, 21,245.85 [ton].

Explanation / Answer

As per stoichiometry for 2moles of A we need 5 moles of B, so per hour we are giving 1 tons

So the moles fed in each hour = 907Kg

Out of this the % of B = 90%

so amount fed of B = 816.3 Kg

So moles of B fed = 816.3 / 60 = 13.6 Kmoles

So moles of A that will react = 13.6 X 2 / 5 moles of A = 5.44 Kmoles

As per stoichiometry the product formed (C) will be = 40 / 5 times of B used = 40 X 13.6 / 5 = 108.8Kmoles

Now out of fed stream 10% is D

so the amount of D fed = 90.7 Kg

Moles of D fed = MAss / mol wt = 90.7 / 24 = 3.78 Kmoles

so moles of E formed = 24 X of moles of D fed = 90.72 Kmoles

The tower can withdraw 0.05Kg

so molecular weight = mass / moles = 0.05 / 90.72 = 5.5 nearly equal to 6.

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