sulphuric acid is produced when Sulphur dioxide reacts with oxygen and water in the presence of a catalyst : `2SO_(2)(g) + O_(2) (g) + 2H_(2)O(l) to2H_(2)SO_(4)`. If 5.6 mol of `SO_(4)` reacts with 4.8 mol of `O_(2)` and a large excess of water , what is the maximum number of moles of `H_(2)SO_(4)` that can be obtained ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2SO_2(g) + O_2(g) + 2H_2O(l) \rightarrow 2H_2SO_4(l) \]

### Step 2: Identify the moles of reactants
From the question, we have:
- Moles of \( SO_2 \) = 5.6 mol
- Moles of \( O_2 \) = 4.8 mol
- Water is in excess, so we do not need to consider it as a limiting reagent.

### Step 3: Determine the stoichiometric ratios
From the balanced equation:
- 2 moles of \( SO_2 \) react with 1 mole of \( O_2 \) to produce 2 moles of \( H_2SO_4 \).

### Step 4: Calculate the required moles of \( O_2 \) for the given moles of \( SO_2 \)
To find out how much \( O_2 \) is required for 5.6 moles of \( SO_2 \):
\[ \text{Required moles of } O_2 = \frac{5.6 \text{ moles } SO_2}{2} = 2.8 \text{ moles } O_2 \]

### Step 5: Compare the available moles of \( O_2 \) with the required moles
We have 4.8 moles of \( O_2 \) available, which is greater than the 2.8 moles required. Therefore, \( SO_2 \) is the limiting reagent.

### Step 6: Calculate the moles of \( H_2SO_4 \) produced from the limiting reagent
From the balanced equation, we see that:
- 2 moles of \( SO_2 \) produce 2 moles of \( H_2SO_4 \).
- Therefore, 1 mole of \( SO_2 \) produces 1 mole of \( H_2SO_4 \).

Since we have 5.6 moles of \( SO_2 \):
\[ \text{Moles of } H_2SO_4 = 5.6 \text{ moles } SO_2 \times \frac{2 \text{ moles } H_2SO_4}{2 \text{ moles } SO_2} = 5.6 \text{ moles } H_2SO_4 \]

### Conclusion
The maximum number of moles of \( H_2SO_4 \) that can be obtained is **5.6 moles**.

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