A common laboratory preparation of O₂ involved the decomposition of hydrogen peroxide, H₂O₂, according to the equation,

2H₂O₂(aq) ---> 2H₂O(l) + O₂(g)

If 240 mL of O₂ at 23 ûC and at 0.965 atm pressure are collected over a sample of water at the same temperature determine the number of moles of O₂ obtained in the reaction.

Solution:

Since the oxygen is collected by displacement of water as the oxygen is collected water is displaced from a text tube initially filled with water.

(see Figure 5.12 on page 196 in your textbook). So as the oxygen gas fills the test tube water escapes in to the vapor phase and mixes with the oxygen in the gas phase.

A Dalton's Law problem!

So the total pressure exerted by the gas above the water in the test tube is the sum of the pressure due to the water in the vapor phase and the oxygen in the vapor phase.

P_T = PO₂ + PH₂O

At 23 ûC the PH₂O is 21.0 mmHg this can be obtained from a table of vapor pressures.

Sine the question asks for the number of mol of oxygen we must eliminate the water in the gas phase above the sample,

PO₂ = P_T - PH₂O = 733 mmHg - 21.0 mmHg = 712 mmHg (.937 atm)

Now that we have the pressure of only oxygen in the test tube we can calculate the mol of oxygen;

nO₂ = PV·(RT)^-1 = (0.937 atm . 0.240 L)· (0.0821 L·atm mol^-1·K^-1 · 296 K)^-1 = 9.25 x 10^-3 mol O₂