Question

A reaction 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?

Answer 1

Answer:

82.08 %

Explanation:

• The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100.

Actual yield = 26.80 kg.

• To get the calculated yield:

• The balanced equation of reacting N2 with H2 to produce NH3 is:

N₂ + 3H₂ → 2NH₃

• It is clear that 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

• N₂ is present in excess and H₂ is the limiting reactant.

• We need to convert the mass of H₂ added (5.79 kg) to moles using the relation:

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

• We can get the no. of moles of NH₃ produced.

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 no. of moles of NH₃ produced = (2880.6 mol)(2.0 mol) / (3.0 mol) = 1920.4 mol.

• We can know get the calculated yield of NH₃ = no. of moles x molar mass = (1920.4 mol) (17.00 g/mol) = 32646.76 g ≅ 32.65 kg.

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

Answer 2

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:

$\% = \frac{m_{e}}{m_{t}} \times 100$  (2)

Where:

$m_{e}$: is the experimental (or actual) mass = 26.8 kg

$m_{t}$: 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

$n_{{H_{2}}_{i}} = \frac{m}{M}$

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:

$n_{H_{2}} = \frac{5790 g}{2.016 g/mol} = 2872 moles$

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:  

$n_{NH_{3}} = \frac{2\: moles \:NH_{3}}{3\: moles \: H_{2}}*n_{{H_{2}}_{i}} =  \frac{2\: moles \:NH_{3}}{3\: moles \: H_{2}}*2872 \:moles\: H_{2}} = 1914.7 \:moles$  

The theoretical mass of NH₃ is:

$m_{NH_{3}} = 1914.7 \: moles*17.031 g/mol = 32609.3 g = 32.6 kg  $

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

$\% = \frac{26.8 kg}{32.6 kg} \times 100 = 82.2 \%$

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

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