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?

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

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hope it ,

bioteacher101