KClO3 22℃ 749 torr 0.650 L 22℃ 21 torr O2

KClO3 22℃ 749 torr 0.650 L 22℃ 21 torr O2

 

 

A sample of solid KClO3 was heated in a test tube

and decomposed according to the following reaction:

KClO3(s) → KCl(s) + O2(g)

The oxygen produced was collected by displacement of water

at 22℃ at a total pressure of 749 torr.

The volume of the gas collected was 0.650 L,

and the vapor pressure of water at 22℃ is 21 torr.

Calculate the partial pressure of O2 in the gas collected

and the mass of KClO3 in the sample that was decomposed.

 

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전체 압력 = O2의 분압 + 수증기의 분압

( 참고 https://ywpop.tistory.com/3490 )

 

O2의 분압 = 749 – 21 = 728 torr

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

생성된(포집된) O2의 몰수를 계산하면,

n = PV / RT

= [(728/760) (0.650)] / [(0.08206) (273.15+22)]

= 0.0257 mol O2

 

 

 

균형 맞춘 반응식

2KClO3(s) → 2KCl(s) + 3O2(g)

 

 

 

KClO3 : O2 = 2 : 3 계수비(= 몰수비) 이므로,

분해된 KClO3의 몰수를 계산하면,

KClO3 : O2 = 2 : 3 = ? mol : 0.0257 mol

? = 2 × 0.0257 / 3 = 0.0171 mol KClO3

 

 

 

KClO3의 몰질량 = 122.55 g/mol

0.0171 mol × (122.55 g/mol) = 2.10 g KClO3