3.50 g N2 H2 300 K 1.00 atm 7.46 L mass percent

3.50 g N2 H2 300 K 1.00 atm 7.46 L mass percent

 

 

A mixture containing nitrogen and hydrogen weighs 3.50 g

and occupies a volume of 7.46 L at 300 K and 1.00 atm.

Calculate the mass percent of these two gases.

Assume ideal-gas behavior.

 

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PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

n = PV / RT

= [(1.00) (7.46)] / [(0.08206) (300)]

= 0.303 mol

 

 

 

N2의 질량 = x 라 두면,

H2의 질량 = 3.50 – x

 

 

 

(x / 28) + (3.50 – x) / 2 = 0.303

 

2x + 28(3.50 – x) = 0.303 × 2 × 28

 

2x – 28x = (0.303 × 2 × 28) – 28(3.50)

 

26x = 81.032

 

x = 81.032 / 26 = 3.1166 g

---> N2의 질량

 

 

 

(3.1166 / 3.50) × 100 = 89.0457%

= 89.0% N2

---> 11.0% H2