Calculating Final Composition of Compounds in Haber's Process with 50% Yield

  • Context: Chemistry 
  • Thread starter Thread starter Suyash Singh
  • Start date Start date
  • Tags Tags
    Concept Mole Moles
Click For Summary
SUMMARY

The final composition of compounds in Haber's process with a 50% yield from 30 litres of N2 and 30 litres of H2 results in 10 litres of NH3, 15 litres of H2, and 25 litres of N2. The stoichiometric equation N2 + 3H2 -> 2NH3 indicates that hydrogen is the limiting reagent. Given the yield definition, only half of the theoretical ammonia is produced, confirming that all hydrogen is consumed while nitrogen remains in excess.

PREREQUISITES
  • Understanding of stoichiometry in chemical reactions
  • Familiarity with the Haber process and its chemical equation
  • Knowledge of yield calculations in chemical processes
  • Basic principles of gas laws and conservation of mass
NEXT STEPS
  • Study stoichiometric calculations in chemical reactions
  • Learn about the Haber process and its industrial applications
  • Research yield calculations and their significance in chemical engineering
  • Explore gas laws and their implications in reaction conditions
USEFUL FOR

Chemistry students, chemical engineers, and anyone interested in understanding the principles of chemical reactions and yield calculations in industrial processes.

Suyash Singh
Messages
168
Reaction score
1

Homework Statement


30 litres of N2 and 30 litres of H2 are taken in Habers process with 50% yield.What is the final composition of the compunds in litres?

Homework Equations


N2+3H2->2NH3

The Attempt at a Solution


N2 + 3H2-> 2NH3
intial: 30 30 0
final: 30 - x/2 30-(3x)/2 (2x)/2

so,
30 - x/2 + 30-(3x)/2 = (2x)/2

this gives x=20
N2=20L
H2=0L
NH3=20L

Am i right?
 
Physics news on Phys.org
Suyash Singh said:
so,
30 - x/2 + 30-(3x)/2 = (2x)/2
Why do you expect these two to be equal?

How is the yield defined? Using all hydrogen with just 50% yield is odd.
 
mfb said:
Why do you expect these two to be equal?

Howis the yield defined? Using all hydrogen with just 50% yield is odd.
litres are conserved
50 percent yield means half of theoretical value of NH3
 
Suyash Singh said:
litres are conserved

2 L of hydrogen plus 1 L of oxygen produce 2 L of gaseous water, 2+1 yields 2, what is conserved here?

Suyash Singh said:
50 percent yield means half of theoretical value of NH3

If only 50% of ammonia was produced, was all the limiting reagent consumed?
 
Borek said:
2 L of hydrogen plus 1 L of oxygen produce 2 L of gaseous water, 2+1 yields 2, what is conserved here?
If only 50% of ammonia was produced, was all the limiting reagent consumed?
Ok i understand the limiting concept now.

NH3 produced=2/3 x 30 =20 L

H2 left= 0 L

N2 left= 30- (1/2 x 20) =20 L
 
So which one is the limiting reagent here? And how much was left if only 50% reacted?
 
  • Like
Likes   Reactions: Suyash Singh
Borek said:
So which one is the limiting reagent here? And how much was left if only 50% reacted?
H2 the limiting reagent

all the elements react only 50 %?
 
Suyash Singh said:
H2 the limiting reagent

Right.

all the elements react only 50 %?

Stop guessing, use stoichiometry to calculate.
 
Suyash Singh said:
50 percent yield means half of theoretical value of NH3
But you got all of the maximum yield of NH3: You used up all hydrogen.
Based on that you can figure out how much NH3 is produced if you use only half of it.
 
  • #10
NH3 produced = 50/100 x (2/3 x 30) = 10 L
H2 left
= 30 - (3/2 x 10) = 15 L
N2 left
= 30 - (1/2 x 10)= 25 L
 
  • #11
That looks good.
 
  • Like
Likes   Reactions: Suyash Singh

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
3K
Replies
2
Views
7K
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
18K
Replies
3
Views
13K