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

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:

(a)

The balance chemical equation for the synthesis of Ammonia is as follow;

N₂ + 3 H₂ → 2 NH₃

Step 1: Calculate moles of N₂ as;

Moles = Mass / M/Mass

Moles = 9.75 g / 28.01 g/mol

Moles = 0.348 moles

Step 2: Calculate moles of H₂ as;

Moles = Mass / M/Mass

Moles = 1.86 g / 2.01 g/mol

Moles = 0.925 moles

Step 3: Find 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 Hydrogen is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculate 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: Calculate theoretical yield of Ammonia as,

Theoretical Yield = Moles * M. Mass

Theoretical Yield = 0.616 mol * 17.03 g/mol

Theoretical Yield = 10.50 g

(b)

%age yield = Actual Yield / Theoretical Yield * 100

%age yield = 2.87 g / 10.50 g * 100

%age yield = 27.33 %