One commercial system removes `SO_(2)` emission from smoke at `95(@)C` by the following set of reaction :
`SO_(2)(g)+Cl_(2)(g) to SO_(2)Cl_(2)`(g)
`SO_(2)Cl_(2)(g) +H_(2)O(l) to H_(2)SO_(4)+HCl`
`H_(2)SO_(4)+Ca(OH)_(2) to CaSO_(4)+H_(2)O`
How many grams of `CaSO_(4)` may be produced from 3.78 g of `SO_(2)` ?

Answer

To determine how many grams of \( \text{CaSO}_4 \) may be produced from 3.78 g of \( \text{SO}_2 \), we will follow these steps:

### Step 1: Calculate the number of moles of \( \text{SO}_2 \)

First, we need to find the molar mass of \( \text{SO}_2 \):
- Sulfur (S) has a molar mass of approximately 32 g/mol.
- Oxygen (O) has a molar mass of approximately 16 g/mol.

Thus, the molar mass of \( \text{SO}_2 \) is:
\[ \text{Molar mass of } \text{SO}_2 = 32 + (2 \times 16) = 32 + 32 = 64 \text{ g/mol} \]

Now, we can calculate the number of moles of \( \text{SO}_2 \) in 3.78 g:
\[ \text{Moles of } \text{SO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{3.78 \text{ g}}{64 \text{ g/mol}} \approx 0.059 \text{ moles} \]

### Step 2: Determine the moles of \( \text{CaSO}_4 \) produced

From the reactions provided:
1. \( \text{SO}_2 + \text{Cl}_2 \rightarrow \text{SO}_2\text{Cl}_2 \) (1:1 ratio)
2. \( \text{SO}_2\text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + \text{HCl} \) (1:1 ratio)
3. \( \text{H}_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} \) (1:1 ratio)

From the stoichiometry of the reactions, we see that 1 mole of \( \text{SO}_2 \) produces 1 mole of \( \text{CaSO}_4 \). Therefore, the moles of \( \text{CaSO}_4 \) produced will also be approximately 0.059 moles.

### Step 3: Calculate the mass of \( \text{CaSO}_4 \)

Next, we need to find the molar mass of \( \text{CaSO}_4 \):
- Calcium (Ca) has a molar mass of approximately 40 g/mol.
- Sulfur (S) has a molar mass of approximately 32 g/mol.
- Oxygen (O) has a molar mass of approximately 16 g/mol.

Thus, the molar mass of \( \text{CaSO}_4 \) is:
\[ \text{Molar mass of } \text{CaSO}_4 = 40 + 32 + (4 \times 16) = 40 + 32 + 64 = 136 \text{ g/mol} \]

Now, we can calculate the mass of \( \text{CaSO}_4 \) produced:
\[ \text{Mass of } \text{CaSO}_4 = \text{moles} \times \text{molar mass} = 0.059 \text{ moles} \times 136 \text{ g/mol} \approx 8.024 \text{ g} \]

### Final Answer
The mass of \( \text{CaSO}_4 \) produced from 3.78 g of \( \text{SO}_2 \) is approximately **8.024 g**.

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