 Chemistry
16 May, 04:59

# 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?

+2
1. 16 May, 05:57
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(a) Theoretical Yield = 10.50 g

(b) %age yield = 27.33 %

Explanation:

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