10 moles of `SO_(2)` and 4 moles of `O_(2)` are mixed in a closed vessel of volume 2 litres. The mixture is heated in the presence of Pt catalyst. Following reaction takes place:
`2SO_(2)(g)+O_(2)(g)to2SO_(3)(g)`
Assuming the reaction proceeds to completion.
Select the correct statement.

Options

A. `SO_(2)` is the limiting reagent
B. `O_(2)` is the limiting reagent
C. Both `SO_(2)` and `O_(2)` are limiting
D. Cannot be predicted

Answer

To solve the problem, we need to determine which reactant is the limiting reagent in the reaction between sulfur dioxide (SO₂) and oxygen (O₂) to form sulfur trioxide (SO₃). Here’s a step-by-step solution:

### Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction is:
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g) \]

### Step 2: Identify the number of moles of each reactant.
From the problem, we have:
- Moles of SO₂ = 10 moles
- Moles of O₂ = 4 moles

### Step 3: Determine the stoichiometric coefficients.
From the balanced equation, we can see the stoichiometric coefficients:
- For SO₂: 2
- For O₂: 1

### Step 4: Calculate the mole ratio for each reactant.
We will calculate the ratio of the moles of each reactant to its stoichiometric coefficient.

**For SO₂:**
\[ \text{Ratio for SO}_2 = \frac{\text{Moles of SO}_2}{\text{Stoichiometric coefficient of SO}_2} = \frac{10}{2} = 5 \]

**For O₂:**
\[ \text{Ratio for O}_2 = \frac{\text{Moles of O}_2}{\text{Stoichiometric coefficient of O}_2} = \frac{4}{1} = 4 \]

### Step 5: Compare the ratios to identify the limiting reagent.
The limiting reagent is the one with the lower ratio:
- Ratio for SO₂ = 5
- Ratio for O₂ = 4

Since 4 (for O₂) is less than 5 (for SO₂), O₂ is the limiting reagent.

### Step 6: Conclusion.
The correct statement is that oxygen (O₂) is the limiting reagent in this reaction.

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