Read the passage given below and answer the following questions:
10 mole SO2 and 4 mole of O2 are mixed in a closed vessel of volume 2L.
The mixture is heated in presence of Platinum catalyst. Following reaction takes place :
2SO2(g) + O2(g) → 2SO3(g)
Assuming the reaction proceeds to completion.
Answer the following questions:
QA Select the correct statement –
(a) SO2 is the limiting reagent.
(b) O2 is the limiting reagent.
(c) Both SO2 and O2 are limiting reagent.
(d) Cannot be predicted.
QB Number of mole of SO3 formed in the reaction will be –
(a) 10 (b) 4 (c) 8 (d) 14
QC Number of mole of excess reagent remaining ---
(a) 4 (b) 2 (c) 6 (d) 8
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Answers
Answer:
answer the following qtn:
1.a. so2 is the limitinh reagent
The Main Answer is: QA) (b) O₂ is the limiting reagent
QB) (d) 14
QC) (b) 2
Given: 2SO₂(g) + O₂(g) → 2SO₃(g)
No. of moles of SO₂ = 10
No. of moles of O₂ = 4
To Find: QA) The limiting reagent
QB) Number of moles of SO₃ formed in the reaction
QC) Number of moles of excess reagent remaining
Solution:
2SO₂(g) + O₂(g) → 2SO₃(g)
QA-
From this reaction, we infer that-
1 mole of O₂ reacts with ⇒ 2 moles of SO₂
∴ 4 moles of O₂ react with ⇒ 8 moles of SO₂
But we have 10 moles of SO₂.
∴ O₂ is the limiting reagent.
QB-
As mentioned above, 4 moles of O₂ will react with 8 moles of SO₂ to form SO₃.
From the reaction-
1 mole of O₂ produces ⇒ 2 moles of SO₃
∴ 4 moles of O₂ produce ⇒ 8 moles of SO₃
QC-
As mentioned above,
only 8 moles of SO₂ react with O₂ to form the product.
But, we have a total of 10 moles of SO₂.
Therefore, there is an excess of 2 moles of SO₂.
Thus, the final answers are:
QA) (b) O₂ is the limiting reagent
QB) (d) 14
QC) (b) 2
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