In the Haber-Bosch process, nitrogen and hydrogen gas are reacted under high temperature 500 degrees C and pressure 200 atm to form ammonia gas (NH3) according to the following equation: N2+3H2 = 2NH3
If 9300 L H2 gas is reacted with an excess of N2 gas at 200 atm and 500 degrees C, how many moles of ammonia gas would be produced, assuming a 15% yield?
To determine the number of moles of ammonia gas produced, we first need to calculate the number of moles of hydrogen gas (H2) that react in the Haber-Bosch process.
We have given the volume of hydrogen gas (H2) as 9300 L. To convert the volume of gas to moles, we can use the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
Given:
P = 200 atm
V = 9300 L
T = 500 °C = 500 + 273.15 = 773.15 K
Using the ideal gas law, we can rearrange it to solve for n (moles):
n = PV / RT
Substituting the values:
n = (200 atm * 9300 L) / (0.0821 L·atm/(mol·K) * 773.15 K)
Simplifying:
n ≈ 2382.34 mol
So, we have approximately 2382.34 moles of hydrogen gas (H2) reacting.
According to the balanced chemical equation for the Haber-Bosch process, one mole of nitrogen gas (N2) reacts with three moles of hydrogen gas (H2) to produce two moles of ammonia gas (NH3).
Thus, the number of moles of ammonia gas produced can be calculated using stoichiometry.
Given the moles of hydrogen gas (H2) = 2382.34 mol
Using the stoichiometric ratio:
mol H2 : mol NH3 = 3 : 2
Calculating the moles of ammonia gas:
moles of NH3 = (2/3) * mol H2
moles of NH3 = (2/3) * 2382.34 mol
moles of NH3 ≈ 1588.23 mol
So, approximately 1588.23 moles of ammonia gas would be produced.
However, there is an additional factor mentioned, which is the yield. Yield refers to the percentage of the theoretical amount of product actually obtained in a reaction.
Here, the yield is given as 15%, which means only 15% of the ammonia gas produced actually collects. To find the actual moles of ammonia gas collected, we multiply the number of moles calculated above by the yield percentage.
Actual moles of NH3 = yield (%) * moles of NH3
Actual moles of NH3 = (15/100) * 1588.23 mol
Actual moles of NH3 ≈ 238.24 mol
Therefore, approximately 238.24 moles of ammonia gas would be produced, assuming a 15% yield.
You work this stoichiometry problem. Here is the link again.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Multiply the moles NH3 produced by 0.15 to correct for 15% yield and not 100% yield.