PLEASE SHOW STEP BY STEP PROCESS AND ANSWER FOR MY CHEMICAL PROCESS QUESTION Amm

Question

PLEASE SHOW STEP BY STEP PROCESS AND ANSWER FOR MY CHEMICAL PROCESS QUESTION

Ammonia is burned to form nitric oxide in the following equation (not balanced yet): NH_3 + O_2 rightarrow NO + H_2O If 20 Ibm of ammonia and 80 Ibm of oxygen are fed to a batch reactor: Determine the excess reactant and the % excess; and If after the reaction, 10 lbm of NO is found in the reactor, determine the final total lb-moles in the reactor, the final product composition in both dry and wet basis, and the fractional conversion of the limiting reactant.

Explanation / Answer

the balanced reaction is 2NH3+2.5O2------->2NO+ 3H2O

molar mass : NH3 =17, O2= 32, no=30 A and H2O=18

As per the reaction, molar ratio of reactants NH3: O2= 2:2.5 = 1:1.25

moles of reactants : NH3= 20/17= 1.18 lb moles and O2= 80/32=2.5

actual ratio of reactants = 1.18: 2.5= 1 :2.11

since oxygen ratio id higher than theoretical requirement, O2 is excess . Assuming all NH3 is consumed, O2 consumed = (2.5/2)*1.18 =1.475 lbmoles

Oxygen supplied = 2.5 lb moles, % excess oxygen = (2.5-1.475)/1.475=69.5%

2.

2NH3+2.5O2------->2NO+ 3H2O

moles of NO= 10/30= 0.33, moles of NH3 remaining = moles of NH3 supplied-moles of NH3 reacted= 1.18-0.33= 0.85, moles of O2 remaining= 2.5-(2.5/2)*0.33=2.0875. This is dry basis

total moles = 0.33+0.85+2.0875 =3.2675

Molar composition : dry basis = NO= 0.33/3.2675 =0.1009, NH3= 0.85/3.2675=0.26 and O2= 2.0875/3.2675=0.6365

wet basis : moles of water additionally formed = (3/2)*0.33= 0.495

total moles of product on wer basis = 3.2675+0.495=3.7625

Molar composition : NO= 0.33/3.7625=0.0877, NH3= 0.85/3.7625=0.2259, O2= 2.0875/3.7625=0.5548 H2O= 0.495/3.7625=0.1315

Moles of limitng reactant reated= 0.33, moles of NH3 supplied = 1.18

% conversion of limiting reactant= 100*0.33/1.18=27.96%

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