Question

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)

Answer 1

Answer:

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