If an excess of nitrogen gas reacts with 25.0 L of hydrogen gas, according to the reaction below, how many L of ammonia will be produced?
N2(g) + 3H2(g) ---> 2NH3(g)
When everything is in the gaseous state, one may use a short cut; i.e., it isn't necessary to convert to mols of what you have, then to mols of what you want, and back to moles of what you want. Go directly from L of what you have to L of what you want.
25.0L H2 gas x (2 moles NH3/3 mole H2) = ? L NH3.
What is the oxidizing agent during discharge
To find out how many liters of ammonia will be produced, we need to use the stoichiometry of the balanced chemical equation.
According to the balanced equation, 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia (NH3).
Step 1: Convert the given volume of hydrogen gas to moles.
To do this, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (which is not given)
V = volume of the gas (given as 25.0 L)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (which is not given)
Since the pressure and temperature are not given, we cannot calculate the number of moles using the ideal gas law. However, we can assume that the gas is at standard temperature and pressure (STP), which is 273.15 K and 1 atm.
At STP, 1 mole of any gas occupies 22.4 L.
Therefore, we can calculate the number of moles of hydrogen gas as:
n = V / 22.4
n = 25.0 L / 22.4 L/mol
n ≈ 1.116 moles
Step 2: Use the stoichiometry of the balanced equation.
Using the stoichiometry of the balanced equation, we know that 3 moles of hydrogen gas react with 2 moles of ammonia. Therefore, the number of moles of ammonia produced is:
(2 moles NH3 / 3 moles H2) * (1.116 moles H2)
≈ 0.744 moles NH3
Step 3: Convert moles of ammonia to liters.
Since we want the answer in liters, we can use the relationship between moles and volume at STP. At STP, 1 mole of any gas occupies 22.4 L.
Therefore, the volume of ammonia produced is:
Volume = moles * 22.4 L/mol
Volume = 0.744 moles * 22.4 L/mol
≈ 16.66 L
So, approximately 16.66 liters of ammonia will be produced.