vol F2 gas 2.31 g CaBr2 CaF2 Br2 gas 8.19 atm 35.0℃

vol F2 gas 2.31 g CaBr2 CaF2 Br2 gas 8.19 atm 35.0℃

 

 

What volume (L) of fluorine gas is required

to react with 2.31 g of calcium bromide

to fom calcium fluoride and bromine gas

at 8.19 atm and 35.0℃?

F2(g) + CaBr2(s) → CaF2(s) + Br2(g)

 

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CaBr2의 몰질량 = 199.89 g/mol 이므로,

2.31 g / (199.89 g/mol) = 0.011556 mol CaBr2

 

 

 

F2(g) + CaBr2(s) → CaF2(s) + Br2(g)

 

F2 : CaBr2 = 1 : 1 계수비(= 몰수비) 이므로,

0.011556 mol CaBr2와 반응하는 F2의 몰수 = 0.011556 mol

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= [(0.011556) (0.08206) (273.15+35.0)] / (8.19)

= 0.0357 L

 

 

 

답: 3.57×10^(-2) L