Question

The volume at STP of pure O2
necessary for the complete combustion of 12 g of magnesium of
atomic weight 24 is:

Answer 1

Answer:

So 1 mole of O2 has volume of 22.4 L So 12 g of Magnesium has ?

Explanation:

Magnesium has molar mass 24. So 12 / 24 has 0.5  mole

So 0.5 x 22.4 = 11.2 L

Answer 2

Answer:

5.6 Litre or 5.4 dm³

Explanation:

We can write the reaction b/w Mg and O₂ as

Mg + 1/2 O₂ --> MgO

So here 1 mole of Mg requires half mole of Oxygen!

Now , according to the question

Given mass of Magnesium is 12g

==> Moles of Mg = 12/24 = 0.5 mole (w/M)

1 mole of Mg ≡ 0.5 mole of O₂

==>

0.5 mole of Mg ≡ 0.25 moles of O₂ (0.5/2)

0.25 moles of O₂ in terms of Volume = 22.4L / 4 = 5.6 L or it can be converted as 5.6 dm³

Hope this helps!

Have a nice day!