10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and oxygen then react to yield NO gas and water vapor, what is the final gas pressure inside the vessel at 23ºC
A) 0.285 atm
B) 1.79 atm
C) 3.50 atm
D) 3.67 atm
E) 6.48 atm

Question

Quizplus · Accepted Answer

1. Write the balanced chemical equation for the reaction:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

2. Convert the masses of ammonia and oxygen to moles using their molar masses:
nNH3 = 10.0 g / 17.031 g/mol = 0.587 mol
nO2 = 6.50 g / 31.999 g/mol = 0.203 mol

3. Determine the limiting reactant:
According to the balanced equation, 4 moles of NH3 react with 5 moles of O2. Therefore, the amount of NO produced is limited by the reactant that is present in the smaller stoichiometric amount. In this case, the limiting reactant is O2 because there are only 0.203 mol of O2, which is less than the amount required to react completely with 0.587 mol of NH3.

4. Calculate the amount of NO produced:
From the stoichiometry of the balanced equation, 5 moles of O2 react to produce 4 moles of NO. Therefore, the amount of NO produced is:
nNO = (4/5) × nO2 = (4/5) × 0.203 mol = 0.162 mol

5. Calculate the amount of H2O produced:
From the stoichiometry of the balanced equation, 4 moles of NH3 react to produce 6 moles of H2O. Therefore, the amount of H2O produced is:
nH2O = (6/4) × nNH3 = (6/4) × 0.587 mol = 0.880 mol

6. Calculate the total number of moles of gas in the vessel after the reaction:
nTotal = nNH3 + nO2 - nNO - nH2O
nTotal = 0.587 mol + 0.203 mol - 0.162 mol - 0.880 mol
nTotal = -0.252 mol (Note: The negative value means that the reaction produced more water vapor than the volume of the vessel can hold.)

7. Calculate the partial pressures of each gas using the ideal gas law:
PV = nRT
For NH3: PNH3 = (nNH3RT)/V = (0.587 mol × 0.0821 L·atm/mol·K × 296 K)/5.50 L = 0.211 atm
For O2: PO2 = (nO2RT)/V = (0.203 mol × 0.0821 L·atm/mol·K × 296 K)/5.50 L = 0.073 atm
For NO: PNO = (nNORT)/V = (0.162 mol × 0.0821 L·atm/mol·K × 296 K)/5.50 L = 0.059 atm
For H2O: PH2O = (nH2ORT)/V = (0.880 mol × 0.0821 L·atm/mol·K × 296 K)/5.50 L = 3.25 atm

8. Calculate the total pressure inside the vessel by adding the partial pressures:
PTotal = PNH3 + PO2 + PNO + PH2O
PTotal = 0.211 atm + 0.073 atm + 0.059 atm + 3.25 atm = 3.59 atm

9. Correct the total pressure for the excess water vapor:
Since the vessel can only hold a maximum volume of 5.50 L, any excess water vapor will condense and occupy a smaller volume than it would occupy at the same pressure if it were in the gas phase. The pressure of this condensed water vapor can be approximately calculated using the vapor pressure of water at 23ºC, which is 0.042 atm. Therefore, the corrected total pressure is:
PTotal,corr = PTotal - PH2O,excess
PH2O,excess = PH2O - Pvap(H2O,23ºC) = 3.25 atm - 0.042 atm = 3.21 atm
PTotal,corr = 3.59 atm - 3.21 atm = 0.38 atm

10. Round the final answer to two significant figures:
The final gas pressure inside the vessel at 23ºC is 0.38 atm, which is closest to choice D (3.67 atm), but none of the given choices is an acceptable answer.

100 G of Gaseous Ammonia and 6