Areaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is charged with 5.79 kg of h2 and excess n2. a total of 26.8 kg of nh3 are produced. what is the percent yield of the reaction?

The percent yield of the reaction between hydrogen and nitrogen to produce ammonia if the mass of H₂ is 5.79 kg and the nitrogen is in excess is 82.2%.

The reaction between nitrogen and hydrogen to produce ammonia is the following:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)   (1)

The percent yield of NH₃ is given by:

 (2)

Where:

: is the experimental (or actual) mass = 26.8 kg

: is the theoretical mass

So, we need to find the theoretical mass.

Initially, we have 5.79 kg of H₂ and an excess of N₂. Knowing that the limiting reactant is H₂, we can calculate the mass of ammonia (theoretical mass).  

First, we need to calculate the number of moles of H₂

Where:

m: is the mass = 5.79 kg = 5790 g

M: is the molar mass = 2.016 g/mol      

Hence, the number of moles of H₂ is:

Now, we can find the number of moles of NH₃ knowing that 3 moles of H₂ reacts with 1 mol of N₂ to produce 2 moles of NH₃ (eq 1), so:  

The theoretical mass of NH₃ is:

Finally, the percent yield of the reaction is (eq 2):  

Therefore, the percent yield of the reaction is 82.2%.

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I hope it helps you!

82.08 %

Explanation:

Actual yield = 26.80 kg.

N₂ + 3H₂ → 2NH₃

n = mass /molar mass = (5790 g) / (2.01 g/mol) = 2880.6 mol.

Using cross multiplication:

3.0 mole of H₂ produce → 2.0 moles off NH₃, from the stichiometry.

2880.6 mol of H₂ produces → ??? moles of NH₃.

∴ The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100 = [(26.8 kg) / (32.65 kg)] x 100 = 82.08 %.

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