1.86 g H2 is allowed to react with 9.75 g N2, producing 2.87g NH3.

A. What is the theoretical yield in grams for this reaction under the given conditions?

B. What is the percent yield for this reaction under the given conditions?

(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 %