127℃ 825 torr NH3 33.6 mL 318℃ 735 torr 수증기 몇 mL

127℃ 825 torr NH3 33.6 mL 318℃ 735 torr 수증기 몇 mL

 

 

127℃, 825 torr에 있는 NH3 33.6 mL가

산소와 다음의 반응식으로 반응하면

318℃, 735 torr 하에서 수증기 몇 mL가 생성되는가?

4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)

 

 

How many mL of water vapor, measured at 318℃ and 735 torr,

are formed when 33.6 mL of NH3 at 825 torr and 127℃

react with oxygen according to the following equation?

4NH3 + 3O2 → 2N2 + 6H2O

 

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PV = nRT

( 참고 https://ywpop.tistory.com/3097 )

 

n = PV / RT

= [(825/760) (33.6/1000)] / [(0.08206) (273.15+127)]

= 0.0011108 ≒ 0.00111 mol NH3 반응

 

 

 

NH3 : H2O = 4 : 6 계수비(= 몰수비) 이므로,

0.00111 mol × (6/4) = 0.00167 mol H2O 생성

 

 

 

V = nRT / P

= [(0.00167) (0.08206) (273.15+318)] / (735/760)

= 0.0838 L

= 83.8 mL

 

 

 

답: 83.8 mL

 

 

 

 

또는

33.6 mL NH3 × (6/4) = 50.4 mL H2O

---> 127℃, 825 torr

 

 

318℃, 735 torr에서의 부피를 계산하면,

PV / T = P’V’ / T’

( 참고 https://ywpop.tistory.com/7091 )

 

V’ = PVT’ / P’T

= [(825) (50.4) (273.15+318)] / [(735) (273.15+127)]

= 83.6 mL