24℃ 835 mmHg Mg 12.0 g H2 gas volume

24℃ 835 mmHg Mg 12.0 g H2 gas volume

 

 

24℃, 835 mmHg에서 Mg 12.0 g이 반응해 생성되는 H2 기체의 부피(L)를 구하시오.

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

 

 

Hydrogen gas can be produced in the laboratory

through the reaction of magnesium metal with hydrochloric acid;

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g).

 

 

What is the volume, in liters, of H2 gas produced

at 24℃ and 835 mmHg, from the reaction of 12.0 g of Mg?

 

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Mg의 몰질량 = 24.31 g/mol

12.0 g / (24.31 g/mol) = 0.4936 mol Mg

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Mg : H2 = 1 : 1 계수비(= 몰수비) 이므로,

0.4936 mol Mg 반응 시

생성되는 H2의 몰수 = 0.4936 mol

 

 

 

PV = nRT

( 참고 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= (0.4936) (0.08206) (273.15 + 24) / (835/760)

= 10.9549 L

 

 

 

답: 11.0 L

 

 

 

 

[키워드] 24℃ 835 mmHg Mg 12.0 g H2 gas vol(L)