What volume of O2 at stp required for oxidation of 1L os S02 at 298k and 1atm
Answers
Given:
Volume of SO2 = 1 L
To Find:
Volume of O2 required at STP to oxidise SO2
Calculation:
- The oxidation of SO2 is given as:
2 SO2 (g) + 2 O2 (g) ----> 2 SO3 (g)
- From the above reaction, 2 moles of O2 is required for the oxidation of 2 moles of SO2.
⇒ 1 moles of O2 is required for the oxidation of 1 moles of SO2.
- At STP, volume of one mole of any gas = 22.4 L
⇒ 22.4 L of O2 is required for the oxidation of 22.4 L of SO2.
⇒ 1 L of O2 is required for the oxidation of 1 L of SO2.
The volume of O₂ at STP required for oxidation of 1 L of SO₂ at 298 K and 1 atm is 1 Liter.
Given:
Volume of SO₂ = 1 L
Step-by-step explanation:
The chemical equation of oxidation of SO₂ is given below:
2 SO₂ (g) + 2 O₂ (g) → 2 SO₃ (g)
From above reaction, we can understand that 2 moles of O₂ is required for the oxidation of 2 moles of SO₂
From which 1 moles of O₂ is required for the oxidation of 1 moles of SO₂
The volume of one mole of any gas at STP = 22.4 L
22.4 L of O₂ is required for oxidation of 22.4 L of SO₂
∴ 1 L of O₂ is required for the oxidation of 1 L of SO₂