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calculate the volume of air containing 21% oxygen by volume at stp required to convert 294 cm3 of sulphur dioxide to sulphur trioxide under the same conditions ?

calculate the volume of air containing 21% oxygen by volume at stp required to convert 294 cm3 of sulphur dioxide to sulphur trioxide under the same conditions ?

Grade:11

1 Answers

Vikas TU
14149 Points
3 years ago
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

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