nitrogen and hydrogen react to form ammonia. N2+3H2=2NH3. 12 dm3 of hydrogen reacted with excess nitrogen to form 2 dm3 of ammonia. what is the percentage yield of ammonia at room temperature and pressure
N2 + 3H2 ==> 2NH3
With gases one can take a shortcut and work the problem as if volume = mols.
12 dm3 H2 x (2 dm3 NH3/3 dm3 H2) = 8 dm3 NH3 if the yield were 100%. It formed 2 dm3; therefore,
%yield = (2/8)*100 = ?%.
To calculate the percentage yield of ammonia, we need to compare the actual yield to the theoretical yield. The balanced equation tells us that 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2) to produce 2 moles of ammonia (NH3).
First, let's calculate the number of moles of hydrogen that reacted using the ideal gas law:
PV = nRT
Where:
P = pressure in atm (room temperature and pressure are typically around 1 atm)
V = volume in liters (12 dm^3 = 12 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (room temperature is approximately 298 K)
Rearranging the equation to solve for n:
n = PV / RT
= (1 atm) * (12 L) / (0.0821 L·atm/(mol·K) * 298 K)
≈ 0.490 mol
According to the balanced equation, 3 moles of hydrogen react to form 2 moles of ammonia. Therefore, the calculated number of moles of ammonia produced would be:
(2/3) * 0.490 mol
≈ 0.327 mol
To convert this to volume, we can use the molar volume of gases, which is approximately 22.4 L/mol at room temperature and pressure:
0.327 mol * 22.4 L/mol
≈ 7.33 L
The actual yield is given as 2 dm^3, which is equivalent to 2 L.
Now we can calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) * 100
= (2 L / 7.33 L) * 100
≈ 27.3%
Therefore, the percentage yield of ammonia at room temperature and pressure is approximately 27.3%.
To calculate the percentage yield of ammonia, you'll need to follow these steps:
Step 1: Determine the stoichiometry of the balanced equation.
According to the balanced equation, 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2) to form 2 moles of ammonia (NH3).
Step 2: Calculate the number of moles of hydrogen that reacted.
Given that you have 12 dm3 of hydrogen, we need to convert this volume into moles. To do this, we use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (atm)
V = volume (dm3)
n = number of moles
R = ideal gas constant (0.0821 dm3⋅atm/mol⋅K)
T = temperature (Kelvin)
Assuming room temperature and pressure (around 298 K and 1 atm), we can use the ideal gas law to calculate the number of moles of hydrogen:
n = PV / RT
= (1 atm) * (12 dm3) / (0.0821 dm3⋅atm/mol⋅K * 298 K)
≈ 0.489 moles of hydrogen
Step 3: Use the stoichiometry of the balanced equation to determine the expected yield of ammonia.
Since 3 moles of hydrogen are required to produce 2 moles of ammonia, the expected number of moles of ammonia can be calculated using a ratio:
(2 moles of NH3 / 3 moles of H2) * (0.489 moles of H2)
≈ 0.326 moles of NH3
Step 4: Calculate the percentage yield of ammonia.
Percentage yield = (Actual yield / Theoretical yield) * 100
In this case, the actual yield of ammonia is given as 2 dm3. To find the actual number of moles, we again use the ideal gas law:
n = PV / RT
= (1 atm) * (2 dm3) / (0.0821 dm3⋅atm/mol⋅K * 298 K)
≈ 0.081 moles of NH3
Now we can calculate the percentage yield:
Percentage yield = (0.081 moles / 0.326 moles) * 100
≈ 24.8%
Therefore, the percentage yield of ammonia at room temperature and pressure is approximately 24.8%.