Ammonia reacts with oxygen at 120 degrees C to form nitrogen monoxide and water in a sealed 40-L container. When 34 g of ammonia reacts with 96 g of oxygen, what is the partial pressure of the nitrogen monoxide? What is the total pressure? Show the balanced reaction.
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34 g NH3 / (17 g/mol) = 2 mol NH3
96 g O2 / (32 g/mol) = 3 mol O2
Our reaction is: 4 NH3 + 5 O2 --> 4 NO + 6 H2O
We need to find our limiting reagent. Given 2 mol of NH3, that would require 2*(5/4) = 2.5 mol O2. Since we have more than that (3 mol O2), NH3 must be our limiting reagent.
During the reaction, we consume all of the NH3 and 2.5 mol of O2. That leave 0.5 mol O2 and produces:
2 mol NO
2*(6/4) = 3 mol H2O
Thus, our total moles remaining in the container is: 0.5 + 2 + 3 = 5.5 mol
So, our mole fraction of NO is: 2 / 5.5 = 0.36364
The total pressure in the container is given by the ideal gas law:
P*(40 L) = (5.5 mol)(0.08206)(120 + 273)
P(total) = 4.434 atm
The partial pressure of NO is: (0.36364)(4.434 atm) = 1.61 atm
96 g O2 / (32 g/mol) = 3 mol O2
Our reaction is: 4 NH3 + 5 O2 --> 4 NO + 6 H2O
We need to find our limiting reagent. Given 2 mol of NH3, that would require 2*(5/4) = 2.5 mol O2. Since we have more than that (3 mol O2), NH3 must be our limiting reagent.
During the reaction, we consume all of the NH3 and 2.5 mol of O2. That leave 0.5 mol O2 and produces:
2 mol NO
2*(6/4) = 3 mol H2O
Thus, our total moles remaining in the container is: 0.5 + 2 + 3 = 5.5 mol
So, our mole fraction of NO is: 2 / 5.5 = 0.36364
The total pressure in the container is given by the ideal gas law:
P*(40 L) = (5.5 mol)(0.08206)(120 + 273)
P(total) = 4.434 atm
The partial pressure of NO is: (0.36364)(4.434 atm) = 1.61 atm