3 H2 (g) + N2 (g) --> 2 NH3 (g)
How many liters of ammonia gas can be produced from 10.0 L of N2 and
20.0 L of H2? Assume all gases are measured at the same temperature and pressure.
This is a limiting reagent problem but since reactants and products are gases you may take a shortcut and use volume as moles.
Using the coefficients in the balanced equation, convert 10.0 L N2 to L NH3. Do the same for 20.0 L H2. The answers will not agree; the correct answer in limiting reagent problems is ALWAYS the smaller value.
To find out how many liters of ammonia gas can be produced from the given volumes of N2 and H2 gas, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation tells us that 3 moles of hydrogen gas combine with 1 mole of nitrogen gas to produce 2 moles of ammonia gas. From this, we can establish a molar ratio between the reactants (H2 and N2) and the product (NH3).
Let's determine the number of moles of N2 and H2 using the ideal gas law:
PV = nRT
Where:
P = pressure (assumed to be constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (assumed to be constant)
Since the temperature and pressure are the same for all gases, we can ignore them in this calculation.
Using the ideal gas law, we can calculate the number of moles for each gas:
Number of moles of N2: n(N2) = V(N2) / Vm(N2)
Number of moles of H2: n(H2) = V(H2) / Vm(H2)
Where:
V(N2) = 10.0 L (Volume of N2)
V(H2) = 20.0 L (Volume of H2)
Vm(N2) = molar volume of N2 (22.4 L/mol)
Vm(H2) = molar volume of H2 (22.4 L/mol)
Substituting the values:
n(N2) = 10.0 L / 22.4 L/mol
n(H2) = 20.0 L / 22.4 L/mol
n(N2) ≈ 0.4464 mol
n(H2) ≈ 0.8929 mol
From the balanced chemical equation, we know that the molar ratio between N2 and NH3 is 1:2. Therefore, the number of moles of NH3 produced will be twice the number of moles of N2 used:
n(NH3) = 2 * n(N2)
n(NH3) ≈ 2 * 0.4464 mol
n(NH3) ≈ 0.8928 mol
Finally, we can calculate the volume of ammonia gas produced using the ideal gas law:
V(NH3) = n(NH3) * Vm(NH3)
Where:
n(NH3) = 0.8928 mol (number of moles of NH3)
Vm(NH3) = molar volume of NH3 (22.4 L/mol)
Substituting the values:
V(NH3) = 0.8928 mol * 22.4 L/mol
V(NH3) ≈ 19.99 L
Therefore, approximately 19.99 liters of ammonia gas can be produced from 10.0 L of N2 and 20.0 L of H2.