A reaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is charged with 6.29 kg of H2 and excess N2. A total of 29.2 kg of NH3 are produced. What is the percent yield of the reaction?

81.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar mass of H2 = 2 ×1 = 2 g/mol

Mass of H2 from the balanced equation = 3 × 2 = 6 g

Converting 6 g of H2 to kg, we have:

1000 g = 1 kg

Therefore,

6 g = 6 g × 1 kg /1000 g

6 g = 0.006 kg

Molar mass of NH3 = 14 + (3×1) = 17 g/mol

Mass of NH3 from the balanced equation = 2 × 17 = 34 g

Converting 34 g of NH3 to kg, we have:

1000 g = 1 kg

Therefore,

34 g = 34 g × 1 kg / 1000 g

34 g = 0.034 Kg

Summary:

From the balanced equation above,

0.006 kg of H2 reacted to produce 0.034 kg of NH3.

Next, we shall determine the theoretical yield of NH3. This can be obtained as follow:

From the balanced equation above,

0.006 kg of H2 reacted to produce 0.034 kg of NH3.

Therefore, 6.29 kg of H2 will react to produce = (6.29 × 0.034)/0.006 = 35.64 Kg of NH3.

Thus, the theoretical yield of NH3 is 35.64 kg.

Finally, we shall determine the percentage yield of NH3 as shown below:

Actual yield of NH3 = 29.2 kg

Theoretical yield of NH3 = 35.64 kg

Percentage yield of NH3 =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 29.2/35.64 × 100

Percentage yield of NH3 = 81.9%