4.0 L of hydrogen gas reacts with 2.0 L of nitrogen gas (at the same temperature and pressure), producing ammonia gas, NH3. Find the limiting reactant.
Write the equation and balance it.
Use L N2 and determine L NH3 produced.
Use L H2 and determine L NH3 produced.
The SMALLER of the two will be the correct one to choose and that reagent will be the limiting reagent.
To find the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced chemical equation.
The balanced chemical equation for the reaction is:
3H2 + N2 -> 2NH3
According to the stoichiometric ratio, for every 3 moles of hydrogen gas, we need 1 mole of nitrogen gas to produce 2 moles of ammonia gas.
To determine the limiting reactant, we will compare the number of moles of hydrogen gas to nitrogen gas, using the ideal gas law equation:
PV = nRT
Where:
P = pressure (constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (constant)
Since the temperature and pressure are the same for both gases, the equation can be simplified to:
V/n = constant
Now, let's calculate the number of moles for each gas:
For hydrogen gas (H2):
V = 4.0 L
n = V/constant
For nitrogen gas (N2):
V = 2.0 L
n = V/constant
Since the constant is the same for both gases (as they are at the same temperature and pressure), we can compare the ratios of the number of moles of each gas to their respective stoichiometric ratios:
For hydrogen gas:
n(H2)/3 = (4.0 L)/constant
For nitrogen gas:
n(N2)/1 = (2.0 L)/constant
Comparing these ratios to the stoichiometric ratio (1/3 for hydrogen gas and 1/1 for nitrogen gas), we can determine which reactant is limiting:
If n(H2)/3 < 1/3, hydrogen gas is the limiting reactant.
If n(N2)/1 < 1/1, nitrogen gas is the limiting reactant.
Therefore, by calculating the ratios, we can determine the limiting reactant.