A 5.00L reaction vessel contains hydrogen at a partial pressure of 0.588atm and oxygen gas at a partial pressure of 0.302atm. The equation is 2H2(g)+O2(g)-> 2H2O(g). Suppose the gas mixture is ignited and the reaction produces the theoretical yield of the product. What would be the partial pressure of each substance present in the final mixture?

figure the moles of each reaction gas from

PV=nRT solve for n in each case.

Then you have a limiting reactants problem moles H2 and Moles O2. In a complete reaction, you should have 2 moles of H2 to each mole of O2. Examine your ration. If it is higher than that, then oxygen is limiting, so you figure the mole of steam based on the moles of oxygen. For each mole of oxygen you consumed, you consumed twice that amount of hydrogen, and made twice that amount of steam.
Now figure the moles of hydrogen left, the amount you started with, minus the amount consumed. So you have moles of hydrogen, oxygen (zero), steam

from these, calculate the pressure of each gas P=nRT/V

NOW, if the ratio of hydrogen to oxygen is less than two, your steam will be based on the moles of hydrogen, and you will have excess oxygen left, Proceed to calcualte the partial pressures as laid out above

To find the partial pressures of each substance present in the final mixture, we will use the mole ratios and the ideal gas law.

First, let's find the number of moles of each gas by using their partial pressures and the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given:
Partial pressure of hydrogen gas (H2) = 0.588 atm
Partial pressure of oxygen gas (O2) = 0.302 atm
Volume (V) = 5.00 L

Ideal gas law equation: PV = nRT

Since we are considering the gases in the reaction vessel, the temperature and volume will remain constant throughout the reaction. So, we can ignore the T and R terms in the equation.

Let's calculate the moles of hydrogen gas (H2):
n(H2) = (P(H2) * V) / (R * T)
= (0.588 atm * 5.00 L) / (0.0821 L·atm/mol·K)
≈ 7.15 mol

Similarly, let's calculate the moles of oxygen gas (O2):
n(O2) = (P(O2) * V) / (R * T)
= (0.302 atm * 5.00 L) / (0.0821 L·atm/mol·K)
≈ 1.84 mol

Using the balanced chemical equation, we can determine the mole ratios between the reactants and products. In this case, the balanced equation is:
2H2(g) + O2(g) -> 2H2O(g)

According to the equation, the mole ratio between H2 and O2 is 2:1. Since we have calculated the number of moles for both hydrogen and oxygen, we can determine that the moles of water (H2O) produced will be twice the number of moles of H2. Therefore, the moles of water (H2O) produced will be:
n(H2O) = 2 * n(H2)
= 2 * 7.15 mol
≈ 14.30 mol

Now, we need to calculate the partial pressure of each substance in the final mixture. Assuming that volume and temperature remain constant, the partial pressure is directly proportional to the number of moles. The moles ratio of the gases remains the same as in the initial reaction mixture.

Partial pressure of H2O (water) = Partial pressure of H2 = 0.588 atm (as there is a 1:1 ratio between H2O and H2)

Partial pressure of O2 = 0.302 atm (as there is no O2 left after all of it reacts)

Therefore, the partial pressure of each substance in the final mixture would be:
Partial pressure of H2O = 0.588 atm
Partial pressure of H2 = 0.588 atm
Partial pressure of O2 = 0.302 atm