Please show step by step I\'m trying to learn these the answers are in black Lim
ID: 493373 • Letter: P
Question
Please show step by step I'm trying to learn these the answers are in black
Limiting reactants, excess reactants, conversion. Ammonia (NH_3) is produced by the reaction of nitrogen (N_2) and hydrogen (H_2) by the reaction N_2 + 3H_2 rightarrow 2NH_3 All reaction components are gases. Suppose that 4,000 kg/hr of N_2 are reacted with 1, 200 kg/hr of H_2, and the conversion of N_2 is 70%. What is the limiting reactant? Show your calculations to substantiate your answer. What is the % excess reactant? What is the conversion of H_2? How much NH_3 is produced in kg/hr? What is the exit gas composition-the mole percent each of N_2, H_2 and NH_3? An engineer has pointed out that we may be able to use air (79% N_2, 21% O_2) in place of N_2. What are the pros and cons of using air in place of N_2 as a reactant?Explanation / Answer
Q1.
From the reaction:
a)
limit reactnat --> the reactant that will finish first (theoretical if 100% is reacted)
so..
mol of N2 = mass/MW = 4000/28 = 142.857 kmol of N2
mol of H2 = mass/MW = 1200 /2 = 600 kmol of H2
ratio is 1:3 so...142.857 kmol of N2 requires 3*142.857 = 428.571 kmol of H2, which we DO have... so NH3 is limiting
b)
%excess reactant --> excess / total * 100% = (600-428.571)/428.571*100% =40.000% is excess of H2
c)
the conversion of H2 --> reacted/total * 100%
since only 70% of N2 is reacting --> 0.7*428.571 = 300 kmol of H2 will react
so --> 300/600 * 100% = 50% is the conversion of H2
d)
NH3 produced at 70% yield:
100 kmol of NH3 --> 1:2 ratio so
100 kmol of NH3 --> 2*100 = 200 kmol of NH3 will be produced
mass = mol*MW = 200 *17 = 3400 kg/h of NH3 are produced
e)
exit gas composition is given:
mol of N2 left --> 30% is left --> 142.857 - 100 = 42.857 kmol of N2 left
H2 left --> 600-300= 300 kmol of H2 left
NH3 produced --> 200 kmol
total kmol = 42.857 +300 +200 = 542.857 kmol of gases
composition
N2 --> 42.857 /542.857 = 0.0789
H2 -->300 /542.857 = 0.55263
NH3 -> 200 /542.857 = 0.3684
f)
Pros --> it is cheaper to use air
cons --> O2 could oxidize H2 or NH3 products to NO, H2O, etc...