4.986 L 25℃ 775 mmHg water vapor 32.6 mmHg gram Ca metal

4.986 L 25℃ 775 mmHg water vapor 32.6 mmHg gram Ca metal

H2의 분압(mmHg)과 반응에 사용된 Ca의 질량(g)

 

 

A sample of calcium metal reacts with

aqueous HCl to yield H2 gas:

Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g)

 

The gas that forms is found to have a volume of 4.986 L

at 25℃ and a pressure of 775 mm Hg.

Assuming that the gas is saturated with water vapor

at a partial pressure of 32.6 mm Hg,

what is the partial pressure in millimeters of mercury of the H2?

How many grams of calcium metal were used in the reaction?

 

 

 

 

수상포집으로 모은 집기병 속의 압력

= H2(g)의 압력 + H2O(g)의 압력

( 참고 https://ywpop.tistory.com/3490 )

 

---> 775 mmHg = H2(g)의 압력 + 32.6 mmHg

 

 

 

수소 기체의 분압을 계산하면,

775 – 32.6 = 742.4 mmHg

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

수소 기체의 몰수를 계산하면,

n = PV / RT

= [(742.4/760) (4.986)] / [(0.08206) (273.15 + 25)]

= 0.19907 mol H2

 

 

 

Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g)

 

Ca : H2 = 1 : 1 계수비(= 몰수비) 이므로,

생성된 H2의 몰수 = 반응한 Ca의 몰수

= 0.19907 mol

 

 

 

Ca의 몰질량 = 40.078 g/mol 이므로,

0.19907 mol × (40.078 g/mol) = 7.98 g Ca

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

답: 742.4 mmHg, 7.98 g

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/18434 ]

Hydrogen gas generated when calcium metal reacts with water is collected.

The volume of gas collected at 30°C and pressure of 988 mmHg is 641 mL.

What is the mass (in grams) of the hydrogen gas obtained?

The pressure of water vapor at 30°C is 31.82 mmHg.

 

 

 

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