850℃ 5.00 atm O2(g) 1.00 mol NH3(g)

850℃ 5.00 atm O2(g) 1.00 mol NH3(g)

 

 

In the first step in the industrial process for making nitric acid,

ammonia reacts with oxygen in the presence of a suitable catalyst

to form nitric oxide and water vapor:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

How many liters of NH3(g) at 850℃ and 5.00 atm are required

to react with 1.00 mol of O2(g) in this reaction?

 

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4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

 

NH3 : O2 = 4 : 5 계수비(= 몰수비) 이므로,

1.00 mol O2와 반응하는 NH3의 몰수를 계산하면,

NH3 : O2 = 4 : 5 = ? mol : 1.00 mol

 

? = 4 × 1.00 / 5 = 0.800 mol NH3

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

0.800 mol NH3의 부피를 계산하면,

V = nRT / P

= (0.800) (0.08206) (273.15 + 850) / (5.00)

= 14.7465 L

 

 

 

답: 14.7 L