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11 July, 15:04

If an enclosure of 0.432 L has a partial pressure of O2 of 3.4*10-6 torr at 28 ∘C, what mass of magnesium will react according to the following equation? 2Mg (s) + O2 (g) →2MgO (s)

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  1. 11 July, 18:27
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    184620169 g.

    We can say that this mass corresponds to 184.6 ton

    Explanation:

    This is the reaction: 2Mg (s) + O₂ (g) → 2MgO (s)

    We have the data of O₂, so let's find out the moles with the Ideal Gases Law.

    First of all, we convert pressure to atm

    760 Torr / 3.4*10⁻⁶ Torr = 223529.4 atm

    P. V = n. R. T

    223529.4 atm. 0.432L = n. 0.082 L. atm/mol. K. 301K

    (223529.4 atm. 0.432L) / (0.082 L. atm/mol. K. 301K) = 3798769 moles

    Gosh!, These are a lot of moles.

    Ratio is 1:2. So with 3798769 moles we would need the double of moles of Mg. → 3798769 mol. 2 = 7597538 moles

    Let's convert this moles into mass (mol. molar mass)

    7597538 mol. 24.3 g/mol = 184620169 g

    We can convert this mass to ton

    1 ton = 1*10⁶ g

    184620169 g / 1*10⁶ g = 184.6 ton
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