How many liters of oxygen are required to react with 2.5 grams of magnesium?

1.16L

Explanation:

First, let us generate the balanced equation for the reaction. This is illustrated below

2Mg + O2 —> 2MgO

Now let us covert 2.5g of Mg given in the question to moles. This is illustrated below:

Molar Mass of Mg = 12g/mol

Mass of Mg = 2.5g

Number of mole of Mg =?

Number of mole = Mass /Molar Mass

Number of mole of Mg = 2.5/24 = 0.104mole

From the equation,

2moles of Mg required 1mole of O2.

Therefore, 0.104mole of Mg will require = 0.104/2 = 0.052mole of O2

Now let us calculate the volume occupy by 0.052mole of O2. This is shown below.

1mole of a gas occupy 22.4L at stp

Therefore, 0.052mole of O2 will occupy = 0.052 x 22.4 = 1.16L

Therefore, 1.16L of O2 is required for the reaction

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