Answer:

The volume of the AIR with 21 % oxygen required to to convert 294cm³ of sulphur dioxide to sulphur trioxide is 700cm³

Explanation:

The equation of the reaction of sulphur dioxide ( SO₂) and Oxygen (O₂) to form sulphur trioxide ( SO₃), is given below:

        2SO₂ (G)    +   O₂ (g)    ⇒     2SO₃ (g)

We can calculate the moles of SO₂ and use it to find the moles of O₂ required by using the equation

We have 294 cm³ of SO₂, we can calculate the moles of the SO₂ by use of Avagadro's LAW

Avagadros law state that 1 mole of any gas occupies exactly 22.4L by volume at STP.

Having this in mind, let us convert 22.4 to litres

1000cm³ = 1 L

Then 294 cm³ = 294cm³/1000

                      = 0.294L

Convert 0.294L to moles:

If 22.4L  = 1 mole

Then 0.294L =  1 mole × 0.294/22.4

                    = 0.013125 moles

Use the mole ratio in the equation to find the moles of the oxygen required:

Mole ratio of SO₂:O₂   is 2:1

Therefore the moles of the oxygen used is

                    0.013125/2 moles = 0.0065625moles

This is the number of moles of oxygen from the equation:

We will use the same Avagadro's law to find the volume of oxygen from the moles

     If 1 mole occupies 22.4 L

Then 0.0065625 moles will occupy ⇒ 0.0065625moles × 22.4/1 mole

                                                         = 0.147 Litre

The amount of oxygen used up in this reaction is 0.147 litres or 147cm³.

The next step is to calculate the amount of air that contains 147cm³ of oxygen ⇒

If 21% = 147cm³

Then 100% = 147 × 100/21

                 = 700cm³

Therefore the volume of air required is 700 cm³