Gas Law Questions.?

Question

The following reaction takes place in a sealed 40.0L container at a temperature of 120 degrees C.

4NH3 + 5O2 ----> NO + 6H2O

When 34.0 g NH3 reacts with 96.0 g O2, what is the partial pressure of NO in the container? (What is the limiting reactant?)

What is the total pressure in the container?

Please just show me the steps because I dont exactly know how to go about this problem.

Answer 1

Let's write this out with the phase identifiers. This is important, because partial pressure only deals with gases:
4NH3 (g) + 5O2 (g) ----> NO (g) + 6H2O (g, because it's over 100C)

Step 1: Find # of moles of reactants.
NH3's molar mass is 17 g/mol (N = 14, H =1)
34.0 g NH3 / 17 g/mol = 2 moles of NH3

O2's molar mass is 32 g/mol (O = 16)
96.0 g O2 / 32 g/mol = 3 moles of O2

The reaction uses a ratio of 4 NH3: 5O2, but you have a ratio of 2:3, so NH3 is the limiting reactant.

Next, find the total pressure:
PV = nRT
P = nRT/V
P = (5 moles of gas * 8.314472 L*kPa / K*mol * 393.15 K) / 40 L
P = 408.6 kPa

Now, we need to find the # of moles of each compound at the end of the reaction.

We have 2 moles of NH3 going in and 3 moles of O2. 2 moles of NH3 will react with 2.5 moles of O2, leaving .5 moles of O2. Since we have 2 moles of NH3 reacting, that means we'll multiply each coefficient in the reaction by .5, so there will be .5 moles of NO, and 3 moles of H2O.

Final mix of gases:
.5 mol O2, .5 mol NO, 3 mol H2O

Finally, to find partial pressure, use the partial pressure equation:
P1 / P = n1 / n
meaning that the partial pressure over total pressure equals the # of moles of one compound over the total # of moles of gas.

P1 / 408.6 kPa = .5 mol NO / 4 mol gas
P1 = 408.6 * .5 / 4
P(NO) = 51.075 kPa