Question

What mass of water vapor is produced from the gas phase reaction of O2 with H2 in a 2.5 L container at 28 °C if the partial pressures of O2 and H2 are PO2 = 0.98 atm and PH2 = 1.24 atm 2 H2 (g)+ O2 (g)→ 2 H2O (g)

Answer 1

mass of water = 2.27 g

Step-by-step explanation:

To do this, let's write again the chemical equation:

2H₂(g) + O₂(g) ------> 2H₂O(g)

Now, we can calculate the mass of water, if we know the number of moles of water. These can be calculated with the moles of Hydrogen, because H2 and H2O have the same coefficient in the reaction.

Now, we have partial pressures, and we can treat these gases as ideal gases, so we can use the ideal gas equation which is:

PV = nRT (1)

From the above expression, we solve for the moles of H2 and O2:

n = PV/RT

We have the temperature of 28 °C (In Kelvin is 28 + 273 = 301 K), the constant R is 0.082 L atm /K mol and the volume is 2.5, so the moles for H2 and O2 are:

nH2 = (1.24 * 2.5) / (0.082 * 301) = 0.1256 moles

nO2 = (0.98 * 2.5) / (0.082 * 301) = 0.0993 moles

Now that we know the moles, let's see who is the limiting reactant. The limiting reactant will be the reactant that is consumed first in the reaction, and the moles consumed would be the moles produced of water so we can calculate the mass of water.

So if: 2 mole H2 --------> 1 mole O2

then: X -----------> 0.0993 moles O2

X = 0.0993 * 2 / 1 = 0.1986 moles of H2 are required.

But we do not have 0.1986 moles of H2, instead we have 0.1256 which means that H2 is the limiting reactant.

In the reaction 2 moles of H2 produces 2 moles of H2O, therefore, 0.1256 moles of H2 produces 0.1256 moles of H2O, so it's mass:

m = n * MM

The reported molar mass of water is 18 g/mol so:

m = 0.1256 * 18

m = 2.27 g

And this is the mass of water vapor produced

Answer 2

Answer:2.27 g H2O

Step-by-step explanation:

Step 1. Get the limiting reagent

1.24 atm of H2 would react completely with 1.24 x (1/2) = 0.62 atm of O2, but there is more O2 present than that (0.98). O2 is in excess and H2 is the limiting reactant.

If the reaction goes to completion:

(1.24 atm H2) x (2 mol H2O / 2 mol H2) = 1.24 atm H2O vapor

n = PV / RT = (1.24 atm) x (2.5 L) / ((0.08205746 L atm/K mol) x (28 + 273)K) = 0.126 mol H2O

(0.126 mol H2O) x (18.01532 g H2O/mol) = 2.27 g H2O