Determine how many liters of nitrogen will be required to produce 87.0 liters of ammonia. The balanced equation for this reaction is: 3H2(g) + N2(g) ==> 2NH3(g).
You can work this as a simple stoichiometry problem. Here is a solved example of one. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
OR, since all are gases, we can do it the short cut way and use ONLY volumes (not convert to moles).
L N2 = L NH3 x (1 mole N2/2 moles NH3) = 87(1/2) = ??
To determine the amount of nitrogen (N2) required to produce a given volume of ammonia (NH3), you need to use the stoichiometry of the balanced equation.
The balanced equation states that 3 moles of hydrogen gas (H2) react with 1 mole of nitrogen gas (N2) to produce 2 moles of ammonia gas (NH3).
First, let's convert the given volume of ammonia (87.0 liters) to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (assume constant)
V = volume of gas (87.0 liters)
n = number of moles of gas (what we want to find)
R = ideal gas constant
T = temperature (assume constant)
Since the pressure and temperature are not given, we will omit them from the equation in this case. Thus, the equation becomes:
V = nR
Now, let's rearrange the equation to solve for the number of moles (n):
n = V / R
Substituting the given volume (87.0 liters) and the ideal gas constant (R), we can calculate the number of moles of ammonia:
n = 87.0 L / 22.414 L/mol
n ≈ 3.88 moles NH3
According to the stoichiometry of the balanced equation, we know that 1 mole of N2 reacts with 2 moles of NH3. Therefore, the number of moles of N2 required would be half of the number of moles of ammonia produced:
n(N2) = 1/2 * n(NH3)
n(N2) ≈ 1/2 * 3.88 moles
n(N2) ≈ 1.94 moles
Finally, to convert the number of moles of nitrogen to liters, we will use the ideal gas law equation again:
V = nRT
V = 1.94 moles * 22.414 L/mol
V ≈ 43.48 liters
Therefore, approximately 43.48 liters of nitrogen will be required to produce 87.0 liters of ammonia.