Step 1: Calculate the number of moles of hydrogen gas (H2)
We can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Assuming standard temperature and pressure (STP), where P = 1 atm and T = 0 °C (273.15 K), we have:
n(H2) = PV/RT = (1 atm)(250.3 L)/(0.08206 L atm/mol K)(273.15 K)
n(H2) = 10.64 mol
Step 2: Determine the limiting reactant
From the balanced chemical equation for the reaction between hydrogen and nitrogen to produce ammonia:
N2 + 3H2 -> 2NH3
We can see that 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2). Therefore, we need to compare the number of moles of hydrogen available (10.64 mol) with the number of moles of nitrogen required (10.64 mol / 3 = 3.55 mol).
Since we have an excess of nitrogen, hydrogen will be the limiting reactant.
Step 3: Calculate the theoretical yield of ammonia (NH3)
From the balanced chemical equation, we know that 3 moles of hydrogen produce 2 moles of ammonia. Therefore, the theoretical yield of ammonia can be calculated as follows:
n(NH3) = (2/3) × n(H2)
n(NH3) = (2/3) × 10.64 mol
n(NH3) = 7.09 mol
Step 4: Calculate the mass of ammonia (NH3)
Finally, we can calculate the mass of ammonia (NH3) produced using its molar mass (17.04 g/mol):
mass(NH3) = n(NH3) × molar mass(NH3)
mass(NH3) = 7.09 mol × 17.04 g/mol
mass(NH3) = 120.8 g
Therefore, the mass of ammonia produced in this reaction is 120.8 grams.
