35℃ 725 torr 16.8 mL CH4 27℃ 654 torr O2 mL

35℃ 725 torr 16.8 mL CH4 27℃ 654 torr O2 mL

 

 

How many mL of O2 measured at 27.0℃ and 654 torr

are needed to react completely with 16.8 mL of CH4

measured at 35.0℃ and 725 torr?

 

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PV = nRT

( 참고 https://ywpop.tistory.com/3097 )

 

n = PV / RT

= [(725/760) (16.8/1000)] / [(0.08206) (273.15 + 35.0)]

= 0.000634 mol CH4

 

 

 

메탄 연소 반응식

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

( 참고 https://ywpop.tistory.com/7441 )

 

CH4 : O2 = 1 : 2 계수비(= 몰수비) 이므로,

필요한 O2의 몰수 = 2 × 0.000634 = 0.001268 mol

 

 

 

V = nRT / P

= (0.001268) (0.08206) (273.15 + 27.0) / (654/760)

= 0.0363 L

 

 

 

답: 36.3 mL