3.31 g Pb(NO3)2 1.62 L 27.0℃ PbO

3.31 g Pb(NO3)2 1.62 L 27.0℃ PbO

 

 

A 3.31 g sample of lead nitrate, Pb(NO3)2, molar mass = 331 g/mol,

is heated in an evacuated cylinder with a volume of 1.62 L.

The salt decomposes when heated, according to the equation:

2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)

Assuming complete decomposition, what is the pressure in the cylinder

after decomposition and cooling to a temperature of 300 K?

Assume the PbO(s) takes up negligible volume.

 

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3.31 g / (331 g/mol) = 0.01 mol Pb(NO3)2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)

 

Pb(NO3)2 : NO2 : O2 = 2 : 4 : 1 계수비(= 몰수비) 이므로,

0.01 mol Pb(NO3)2 분해 시,

생성되는 기체의 몰수를 계산하면,

> NO2의 몰수 = 2 × 0.01 = 0.02 mol

> O2의 몰수 = 0.5 × 0.01 = 0.005 mol

---> 0.02 + 0.005 = 0.025 mol

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

P = nRT / V

= (0.025) (0.08206) (300) / (1.62)

= 0.3799 atm

 

 

 

답: 0.380 atm