Answer:
(a) Theoretical Yield = 10.50 g
(b) %age yield = 27.33 %
Explanation:
Answer-Part-(a)
The balance chemical equation for the synthesis of Ammonia is as follow;
N₂ + 3 H₂ → 2 NH₃
Step 1: Calculating moles of N₂ as;
Moles = Mass / M/Mass
Moles = 9.75 g / 28.01 g/mol
Moles = 0.348 moles of N₂
Step 2: Calculating moles of H₂ as;
Moles = Mass / M/Mass
Moles = 1.86 g / 2.01 g/mol
Moles = 0.925 moles
Step 3: Finding Limiting reagent as;
According to equation,
1 mole of N₂ reacts with = 3 moles of H₂
So,
0.348 moles of N₂ will react with = X moles of H₂
Solving for X,
X = 3 mol × 0.348 mol / 1 mol
X = 1.044 mol of H₂
It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen is limiting reagent. Therefore, H₂ will control the final yield.
Step 4: Calculating moles of Ammonia as,
According to equation,
3 mole of H₂ produces = 2 moles of NH₃
So,
0.925 moles of H₂ will produce = X moles of NH₃
Solving for X,
X = 2 mol × 0.925 mol / 3 mol
X = 0.616 mol of NH₃
Step 5: Calculating theoretical yield of Ammonia as,
Theoretical Yield = Moles × M.Mass
Theoretical Yield = 0.616 mol × 17.03 g/mol
Theoretical Yield = 10.50 g
Answer-Part-(b)
%age yield = Actual Yield / Theoretical Yield × 100
%age yield = 2.87 g / 10.50 g × 100
%age yield = 27.33 %
