Question

When solid KClO3 is heated, it decomposes to give solid KCl and O2 gas. A volume of 262 mL of gas is collected over water at a total pressure 730 mmHg of and 24 ∘C. The vapor pressure of water at 24 ∘C is 22 mmHg:
2KClO3(s)→2KCl(s)+3O2(g)
Part A What was the partial pressure of the O2 gas?
Part B How many moles of O2 gas were in the gas sample?

Answer 1

Answer:

Part A: 708 mmHg

Part B: 0.01 mol O2

Explanation:

Total pressure in a gas mixture = Sum of each partial pressure in the mixture so:

Total pressure = 730mmHg

The mixture has 2 compounds, the O2 and the vapor of water.

730mmHg - 22mmHg = 708 mmHg. Now that we have Pp O2, let's apply the Ideal Gas Law to find the mols

P.V = n . R . T

First of all, covert the mmHg in atm

760 mmHg ____ 1 atm

708 mmHg _____ 708/760 = 0.93 atm

and convert 262mL in L

262/1000 = 0.262L

0.93 atm . 0.262L = n . 0.082 . 297K

(0.93 . 0.262)/ (0.082. 297) = n

0.01 mol = n