what volume of air at n.t.p will be required to oxidise 120ml of sulpher di oxide at n.t.p to sulpher tri oxide if the air contains 21 of oxygen

To determine the volume of air needed to oxidize 120 mL of sulfur dioxide (SO₂) to sulfur trioxide (SO₃) at normal temperature and pressure (NTP), we first need to understand the chemical reaction involved and the stoichiometry of the gases. The reaction can be represented as follows:

The Chemical Reaction

The balanced equation for the oxidation of sulfur dioxide to sulfur trioxide is:

2 SO₂(g) + O₂(g) → 2 SO₃(g)

This equation tells us that two moles of sulfur dioxide react with one mole of oxygen to produce two moles of sulfur trioxide. From this, we can derive the stoichiometric relationships needed for our calculations.

Step-by-Step Calculation

• Volume of SO₂: We start with 120 mL of SO₂.
• Stoichiometry of the Reaction: According to the balanced equation, 2 volumes of SO₂ require 1 volume of O₂. Therefore, for 120 mL of SO₂, we can calculate the required volume of O₂.

Using the ratio from the balanced equation:

Volume of O₂ required = (120 mL SO₂) × (1 mL O₂ / 2 mL SO₂) = 60 mL O₂

Calculating the Volume of Air

Now, we need to find out how much air is needed to provide this volume of oxygen. Since air contains approximately 21% oxygen, we can set up the following calculation:

• Percentage of Oxygen in Air: 21% means that in every 100 mL of air, there are 21 mL of O₂.
• Volume of Air Required: To find the volume of air needed to obtain 60 mL of O₂, we can use the proportion:

Volume of air = (60 mL O₂) × (100 mL air / 21 mL O₂) ≈ 285.71 mL air

Final Result

Therefore, to oxidize 120 mL of sulfur dioxide to sulfur trioxide at NTP, you will require approximately 285.71 mL of air. This calculation illustrates the importance of understanding gas volumes and stoichiometry in chemical reactions, especially in processes involving gases where the volumes can be directly related to the number of moles involved.