Question

A sample of soilid potassium chlorate was heated and decomposed into solid potassium chloride and oxygen gas. The oxygen produced was collected over water at 22 degrees celcius at a total pressure of 745 Torr. the volume of the gas collected was 0.650 L, and the vapor pressure of water at 22 degrees celcius is 21 torr.

2 KClO3(s)=2KCl(s)+3O2(g)

First calculate the partial pressre of oxygen in the gas collected and determine the mass of potassium chlorate in the sample that decomposed.

0.0256 g KClO3

2.32 g KClO3

3.15 g KClO3

2.09 g KClO3

4.73 g KClO3

1.67 g KClO3

Answer 1

p(O2) = total Pressure - p(H2O)

= 745 torr - 21 torr

= 724 torr

we have:

P = 724.0 torr

= (724.0/760) atm

= 0.9526 atm

V = 0.65 L

T = 22.0 oC

= (22.0+273) K

= 295 K

find number of moles using:

P * V = n*R*T

0.9526 atm * 0.65 L = n * 0.0821 atm.L/mol.K * 295 K

n = 2.557*10^-2 mol

this is mol of O2 formed

From reaction,

moles of KClO3 reacted = (2/3)* mol of O2 formed

= (2/3)*2.557*10^-2 mol

= 0.01705 mol

Molar mass of KClO3 = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

we have below equation to be used:

mass of KClO3,

m = number of mol * molar mass

= 1.705*10^-2 mol * 122.55 g/mol

= 2.09 g

Answer: 2.09 g KClO3