One commercial system removes `SO_(2)` emmission from smoke at
`95^(@)`C by the following set of reactions-
`SO_(2)(g)+ Cl_(2)(g) to SO_(2)Cl_(2)(g)`
`SO_(2)Cl_(2) + 2H_(2)O to H_(2)SO_(4) + 2HCl`
`H_(2)SO_(4) + Ca(OH)_(2) to CaSO_(4) + 2H_(2)O`
Assuming the process to be 95% efficient . How many moles
of `CaSO_(4)` may be produced from 128 g `SO_(2).[Ca=40,S-32,O-16]`

Options

A. 1.9 moles
B. 2 mol
C. 3.8 mol
D. 0.95 mol

Answer

To solve the problem step by step, let's follow the outlined reactions and apply the necessary calculations.

### Step 1: Calculate the number of moles of SO₂ from the given mass.
We are given 128 g of SO₂. First, we need to find the molar mass of SO₂.

- Molar mass of SO₂ = Mass of S + 2 × Mass of O
- Molar mass of SO₂ = 32 g/mol (for S) + 2 × 16 g/mol (for O) = 32 + 32 = 64 g/mol

Now, we can calculate the number of moles of SO₂:

\[ \text{Number of moles of SO₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{128 \text{ g}}{64 \text{ g/mol}} = 2 \text{ moles} \]

### Step 2: Determine the moles of CaSO₄ produced.
From the reactions provided, we can see that 1 mole of SO₂ produces 1 mole of CaSO₄. Therefore, if we start with 2 moles of SO₂, we can potentially produce 2 moles of CaSO₄.

### Step 3: Account for the efficiency of the process.
The problem states that the process is 95% efficient. Therefore, the actual amount of CaSO₄ produced will be 95% of the theoretical yield.

\[ \text{Moles of CaSO₄ produced} = \text{Theoretical moles} \times \text{Efficiency} \]
\[ \text{Moles of CaSO₄ produced} = 2 \text{ moles} \times 0.95 = 1.9 \text{ moles} \]

### Final Answer:
The number of moles of CaSO₄ that may be produced from 128 g of SO₂, assuming the process is 95% efficient, is **1.9 moles**.

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