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)

Question

Chemistry

Ray

11 May, 20:15

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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Answers (1)

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Alberto · Answer 1

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