Answer:
1.16L
Step-by-step 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