Question

Consider the reaction: 2KClO3 → 2KCl + 3O2
A sample of 5.45 L of oxygen is collected over water at a total pressure of
735.5 torr at 25°C. How many grams of oxygen have been collected? (at
25°C, the vapour pressure of water is 23.8 torr)

Answer 1

According to the Dalton's Law of Partial Pressures, the total pressure in the container must be the sum of the pressures of the gas one collected and the water vapour:

$P_T = P_w + P_{O_2}$ .

Therefore, the pressure of oxygen is:

$P_{O_2} = P_T-P_w = 735.5-23.8 = 711.7$ torr.

Using the ideal gas law (ideal gas constant $R = 62.3637$ L torr mol^-1 K^-1), the number of the moles of oxygen is:

$n = \frac{pV}{RT} = \frac{711.7\cdot5.45}{62.3637\cdot(273.15+25)} = 0.2086$ mol.

The mass of oxygen collected is then the product of its number of the moles and its molar mass $M = 32.00$ g/mol:

$m = nM = 0.2086\cdot32.00 = 6.675$ g, or 6.68 g.

Answer: 6.68 g of oxygen have been collected.