A sample of solid potassium chlorate was heated and decomposed into solid potass

Question

A sample of solid potassium chlorate was heated and decomposed into solid potassium chloride and oxygen gas. The oxygen produced was collected over water at 22 °C at a total pressure of 745 torr. The volume of the gas collected was 0.650L and the vapor pressure of water at 22 °C is 21 torr. 2 KCIO3(s) 2 KCl(s) + 3 O2(g) First, calculate the partial pressure of oxygen in the gas collected and determine the mass of potassium chlorate in the sample that decomposed. 0.0256 g KCIO3 2.32 g KCIO3 3.15 g KCIO3 2.09 g KCIO3 4.73 g KCIO3 1.67 g KCIO3

Explanation / Answer

we have for O2:

pressure of O2 = total pressure - pressure of water

P = 745 - 21

P = 724 torr

= (724/760) 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:

partial pressure of O2 = 724 torr

2.09 g KClO3

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