765 torr 225℃ CuO 35.5 g H2 vol

765 torr 225℃ CuO 35.5 g H2 vol

 

 

765 torr, 225℃에서

산화구리(II) 35.5 g을 환원시키는데 필요한 H2의 부피

 

 

What volume of H2 gas at 765 torr and 225℃

is needed to reduce 35.5 g of copper(II) oxide

to form pure copper and water?

 

CuO(s) + H2(g) → Cu(s) + H2O(g)

 

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CuO의 몰질량 = 79.55 g/mol

35.5 g / (79.55 g/mol) = 0.446 mol CuO

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

CuO + H2 → Cu + H2O

 

CuO : H2 = 1 : 1 계수비(= 몰수비) 이므로,

필요한 H2의 몰수 = 0.446 mol

 

 

 

PV = nRT

( 참고 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= (0.446) (0.08206) (273.15 + 225) / (765/760)

= 18.1 L

 

 

 

답: 18.1 L