30℃ 5.67 L O2 7.066 mPa Mg mass

Magnesium can be used as a “getter” in evacuated enclosures

to react with the last traces of oxygen.

(The magnesium is usually heated by passing an electric current

through a wire or ribbon of the metal.)

If an enclosure of 5.67 L has a partial pressure of O2 of 7.066 mPa at 30℃,

what mass of magnesium will react according to the following equation?

2Mg(s) + O2(g) → 2MgO(s)

30℃에서 5.67 L 진공관에서

O2의 부분 압력이 7.066 mPa일 때,

반응하는 Mg의 질량은 얼마인가?

1 atm = 101325 Pa 이므로,

(7.066 mPa) (1 Pa / 1000 mPa) (1 atm / 101325 Pa)

= (7.066) (1 / 1000) (1 / 101325)

= 6.9736×10^(-8) atm

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

O2의 몰수를 계산하면,

n = PV / RT

= [(6.9736×10^(-8)) (5.67)] / [(0.08206) (273.15 + 30)]

= 1.589465×10^(-8) mol O2

2Mg(s) + O2(g) → 2MgO(s) 로부터,

반응하는 Mg의 몰수를 계산하면,

Mg : O2 = 2 : 1 = ? mol : 1.589465×10^(-8) mol

? = 2 × (1.589465×10^(-8))

= 3.17893×10^(-8) mol Mg

Mg의 몰질량 = 24.305 g/mol 이므로,

(3.17893×10^(-8) mol) (24.305 g/mol)

= 7.726389365×10^(-7) g Mg

답: 7.73×10^(-7) g

[공식] 반응하는 Mg의 질량(g)

= [(atm) (L)] / [(0.08206) (273.15 + ℃)] × 48.61

= [(6.9736×10^(-8)) (5.67)] / [(0.08206) (273.15 + 30)] × 48.61

= 7.72639×10^(-7) g

[예제] If an enclosure of 0.452 L has a partial pressure of O2 of 3.5×10^(-6) torr at 27℃, what mass of magnesium will react according to the following equation?

[(3.5×10^(-6) / 760) (0.452)] / [(0.08206) (273.15 + 27)] × 48.61

= 4.108×10^(-9) g

[키워드] 화학량론 기준, 2Mg + O2 → 2MgO 반응 기준, getter dic

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30℃ 5.67 L O2 7.066 mPa Mg mass