Ammonia reacts with oxygen at 120 C to form NO and H20 in a sealed 40-L container. When 34 g of ammonia reacts with 96 g of oxygen, what is the partial pressure of the NO? What is the total pressure?

Write and balance the equation.

This is a limiting reagent problem.
Convert 34 g NH3 to mols. mols = g/molar mass.
Convert 96 g O2 to mols.
Using the coefficients in the balanced equation, convert mols NH3 to mols NO.
Do the same for O2.
It is likely that you will obtain two different answers; the correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
Using the limiting reagent value do the following:
a. Calculate mols H2O formed.
b. Calculate mols of the non-limiting reagent used.
c. Subtract mols non-limiting reagent initially less mols used to determine mols remaining unreacted.
Then use PV = nRT to solve for pressure of each gas at the conditions listed. Add all of them to find the total pressure.
Post your work if you get stuck.

To determine the partial pressure of NO and the total pressure, we need to follow certain steps:

Step 1: Convert the given masses of ammonia and oxygen to moles.
Given:
Mass of ammonia (NH3) = 34 g
Molecular mass of ammonia (NH3) = 17 g/mol
Mass of oxygen (O2) = 96 g
Molecular mass of oxygen (O2) = 32 g/mol

To convert the masses to moles, we use the formula:
Number of moles = Given mass / Molecular mass

For ammonia (NH3):
Number of moles of NH3 = 34 g / 17 g/mol
Number of moles of NH3 = 2 moles

For oxygen (O2):
Number of moles of O2 = 96 g / 32 g/mol
Number of moles of O2 = 3 moles

Step 2: Determine the stoichiometry of the reaction.
From the balanced chemical equation, we know that:
NH3 + 5/4 O2 → 1 NO + 3/2 H2O

This means that 1 mole of ammonia reacts with 5/4 moles of oxygen to form 1 mole of NO and 3/2 moles of H2O.

Step 3: Calculate the limiting reactant.
To determine the limiting reactant, we compare the moles of reactants based on the stoichiometry of the reaction.

From the stoichiometry, we can see that for every 2 moles of ammonia, we need (5/4) * 2 = 5/2 moles of oxygen. However, we only have 3 moles of oxygen, which is less than what is required.

Therefore, oxygen is the limiting reactant because it is completely consumed, while ammonia is in excess.

Step 4: Determine the moles of NO produced.
Since oxygen is the limiting reactant and we have used 3 moles of it, according to the stoichiometry, we can conclude that 3 moles of NO are produced.

Step 5: Calculate the partial pressure of NO.
Partial pressure is the pressure exerted by a single gas component in a mixture.

Given:
Volume of the container (V) = 40 L

We know that the partial pressure of a gas is given by the formula:
Partial pressure = (Number of moles / Total moles) * Total pressure

Number of moles of NO = 3 moles (as determined in Step 4)

Total moles of gases = Moles of NH3 + Moles of O2
Total moles = 2 moles (NH3) + 3 moles (O2)
Total moles = 5 moles

Now, we need to calculate the total pressure.

Step 6: Calculate the total pressure.
The total pressure can be calculated using the ideal gas law:
PV = nRT

Where:
P = Total pressure
V = Volume
n = Total moles of gases
R = Ideal gas constant (0.0821 L.atm/(mol.K))
T = Temperature (in Kelvin)

Given:
V = 40 L
n = 5 moles (as calculated above)
R = 0.0821 L.atm/(mol.K)
T = 120 °C = (120 + 273) K = 393 K

Now, we can calculate the total pressure (P) using the ideal gas law:
P * 40 = 5 * 0.0821 * 393

Simplifying the equation:
40P = 162.4935

Dividing both sides by 40:
P = 162.4935 / 40
P = 4.0623 atm

Therefore, the partial pressure of NO is 3/5 * 4.0623 = 2.4374 atm, and the total pressure is 4.0623 atm.