Final answer:
The percent yield of NH₃ produced when 4.600 g of N₂ reacts is calculated to be 38.78%, using the theoretical yield determined from stoichiometric calculations.
Step-by-step explanation:
To determine the percent yield of the reaction N₂(g) + 3H₂(g) → 2NH₃(g), we use the actual yield and theoretical yield. First, we need to find the theoretical yield using stoichiometry.
According to the balanced reaction, 1 mole of nitrogen gas (N₂) produces 2 moles of ammonia (NH₃).
The molar mass of nitrogen gas (N₂) is approximately 28 grams/mole, and for ammonia (NH₃), it is approximately 17 grams/mole.
Thus, 4.600 grams of N₂ (4.600 g / 28 g/mole) gives 0.1643 moles of N₂ which theoretically produces 0.3286 moles of NH₃.
This amount of moles of NH₃ has a mass of (0.3286 moles * 17 g/mole) = 5.5862 grams.
This is the theoretical yield. The actual yield is given as 2.165 grams of NH₃. The percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100 percent:
Percent Yield = (Actual Yield / Theoretical Yield) * 100%
= (2.165 g / 5.5862 g) * 100%
= 38.78%
Therefore, the percent yield of the process is 38.78%.