Question

answer the question.

Answer 1

The Balanced Equation of this Reaction is :

✿  2SO₂ + O₂ =====> 2SO₃

⇒ 2 Moles of SO₂ reacts with One Mole of O₂ to form 2 Moles of SO₃

Multiplying the Whole Equation with 2.5, We get :

⇒ (2.5 × 2) Moles of SO₂ reacts with (2.5 × 1) Moles of O₂ to form (2.5 × 2) Moles of SO₃

⇒ 5 Moles of SO₂ reacts with 2.5 Moles of O₂ to form 5 Moles of SO₃

But, It is given in the Question that : 5 Moles of SO₂ and 5 Moles of O₂ React with Each other to form SO₃ and We found from the Balanced Reaction that Only 2.5 Moles of O₂ is sufficient enough to react with 5 Moles of SO₂

⇒ The Excess Reagent is O₂ , Because Oxygen is not Completely used up in the Reaction

⇒ Unreacted O₂ = (5 Moles - 2.5 Moles) = 2.5 Moles

We know that : One Mole of O₂ Weighs 32 grams

⇒ 2.5 Mole of O₂ Weighs : (2.5 × 32) = 80 grams

⇒ Weight of Excess Reagent Left = 80 grams

⇒ The Limiting Reagent is SO₂ , Because SO₂ is Completely used up in the Reaction and Limits the Reaction

From Reaction : Amount of SO₃ Formed is Equal to 5 Moles

We know that : One Mole of SO₃ Weighs : (32 + 48) = 80 grams

⇒ 5 Moles of SO₃ Weighs : (5 × 80) = 400 grams

⇒ Amount of SO₃ formed = 400 grams