An intermediate reaction used in the production of nitrogen containing fertilizers is that b/w NH3 and O2
4NH3(g) + 5O2(g)--> 4NO(g) + 6H2O(g)
A 150.0-L reaction chamber is charged with reactants to the following partial pressures at 500 degrees C: P(sub NH3)= 1.3 atm, P(sub O2)= 1.5 atm. What is the limiting reactant?
I'm having trouble getting to moles of my reactants from the information given. Can I assume that both gases have a volume of 150.0-L since gases assume the volume of the container? Or do I need to get the moles in a different way? I tried to use
V= nRT/P for each gas to get the volumes, but the number of moles is unknown.
P is the partial pressure given.
V is 150 L for both gases.
n can be calculated.
R is 0.08206 L*atm/mol*K
T is 273+500.
You are correct that gases assume the volume of the container, so in this case, you can assume that both NH3 and O2 have a volume of 150.0 L. However, you cannot directly use the Ideal Gas Law equation, V = nRT/P, to calculate the number of moles because the equation is missing the value of n, which represents the number of moles.
To calculate the moles of NH3 and O2, you need to rearrange the Ideal Gas Law equation to solve for n:
n = PV/RT
For NH3:
P(NH3) = 1.3 atm (given)
V(NH3) = 150.0 L (assumed)
R = 0.08206 L*atm/mol*K (gas constant)
T = 273 + 500 = 773 K (temperature in Kelvin)
Plugging in the values:
n(NH3) = (1.3 atm)*(150.0 L) / (0.08206 L*atm/mol*K)*(773 K)
Similarly, you can calculate the moles of O2 using the given information:
P(O2) = 1.5 atm (given)
V(O2) = 150.0 L (assumed)
n(O2) = (1.5 atm)*(150.0 L) / (0.08206 L*atm/mol*K)*(773 K)
By calculating the moles of NH3 and O2, you can determine which reactant is the limiting reactant by comparing the stoichiometric ratio of the balanced chemical equation. In this case, you need to compare the ratio of the coefficients of NH3 and O2 in the equation:
4 NH3 : 5 O2
Whichever reactant has a smaller number of moles relative to its stoichiometric coefficient will be the limiting reactant.