Consider the balanced equation. N2 + 3H2 Right arrow. 2NH3 What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3? Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100.. 17.8% 33.3% 35.7% 53.4%

53.4%

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 × 1.01 = 2.02 g/mol

Mass of H2 from the balanced equation = 3 × 2.02 = 6.06 g

Molar mass of NH3 = 14 + (3×1.01)

= 14 + 3.03

= 17.03 g/mol

Mass of NH3 from the balanced equation = 2 × 17.03 = 34.06 g

Summary:

From the balanced equation above,

6.06 g of H2 reacted to produce 34.06 g of NH3.

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

From the balanced equation above,

6.06 g of H2 reacted to produce 34.06 g of NH3.

Therefore, 26.3 g of H2 will react to produce = (26.3 × 34.06)/6.06 = 147.82 g of NH3.

Thus, the theoretical yield of NH3 is 147.82 g

Finally, we shall determine the percentage yield of NH3. This is illustrated below:

Actual yield = 79 g

Theoretical yield = 147.82 g

Percentage yield =.?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 79/147.82 × 100

Percentage yield of NH3 = 53.4%

it would be a gas formation

it is a solid

explanation: