Ammonia is often formed by reacting nitrogen and hydrogen gases. How many liters of ammonia gas can be formed from 17.2 L of hydrogen gas at 93.0°C and a pressure of 45.8 kPa??
how many moles in 17.2L?
N2+3H2 = 2NH3
2/3 that many moles of NH3
Then recall the PV = nRT
To determine the volume of ammonia gas formed, we need to use the ideal gas law equation: PV = nRT.
Where:
P = pressure (in Pascal)
V = volume (in cubic meters)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given values to the appropriate units:
Hydrogen gas volume = 17.2 L = 0.0172 m^3
Temperature = 93.0°C = (93 + 273) K = 366 K
Pressure = 45.8 kPa = 45,800 Pa
Now, we need to determine the number of moles of hydrogen gas. To do this, we can use the ideal gas law equation, rearranged to solve for n:
n = (PV) / (RT)
Substituting the given values:
n = (45,800 Pa * 0.0172 m^3) / (8.314 J/(mol·K) * 366 K)
n = 3.5084 mol
According to the balanced chemical equation, one mole of hydrogen gas reacts with one mole of ammonia gas. Therefore, the number of moles of ammonia gas formed will also be 3.5084 mol.
Finally, we can calculate the volume of ammonia gas formed using the ideal gas law equation:
V = (nRT) / P
Substituting the values:
V = (3.5084 mol * 8.314 J/(mol·K) * 366 K) / 45,800 Pa
V = 0.0282 m^3
Therefore, 0.0282 m^3 or 28.2 liters of ammonia gas can be formed from 17.2 liters of hydrogen gas at 93.0°C and a pressure of 45.8 kPa.