Chemistry, asked by adityamahale2003, 11 months ago

Calculate the volume of air containing 21% oxygen by volume at STP, required in order to convert 294cm³ of sulphur dioxide to sulphur trioxide under the same conditions(FIITJEE Question).

Answers

Answered by santy2

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

Answered by shubham424380

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