`FeSO_(4)` undergoes decomposition as
`2FeDO_(4)(s) rarr Fe_(2)O_(3)(s)+ SO_(2)(g) + SO_(3)(g)`
At 1 atm 273 K. If (7.6 gm) `FeSO_(4)` so taken then:
The average molar mass of the gaseous mixture,

Options

A. 72
B. 36
C. 48
D. 60

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the Molar Mass of `FeSO4`
First, we need to calculate the molar mass of iron(II) sulfate (`FeSO4`).

- Molar mass of Fe = 55.85 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (4 oxygen atoms)

Calculating the molar mass of `FeSO4`:
\[ \text{Molar mass of } FeSO4 = 55.85 + 32.07 + (4 \times 16.00) = 55.85 + 32.07 + 64.00 = 151.92 \text{ g/mol} \]

### Step 2: Calculate the Moles of `FeSO4`
Next, we calculate the number of moles of `FeSO4` present in 7.6 grams.

Using the formula:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Moles of } FeSO4 = \frac{7.6 \text{ g}}{151.92 \text{ g/mol}} \approx 0.050 \text{ moles} \]

### Step 3: Determine the Stoichiometry of the Reaction
From the balanced decomposition reaction:
\[ 2 FeSO4 \rightarrow Fe2O3 + SO2 + SO3 \]
We see that 2 moles of `FeSO4` produce 1 mole of `Fe2O3`, 1 mole of `SO2`, and 1 mole of `SO3`.

This means:
- For every 2 moles of `FeSO4`, we get 2 moles of gas (1 mole of `SO2` and 1 mole of `SO3`).

### Step 4: Calculate the Moles of Gases Produced
From the moles of `FeSO4` calculated:
\[ \text{Moles of gases} = 0.050 \text{ moles of } FeSO4 \times \frac{2 \text{ moles of gases}}{2 \text{ moles of } FeSO4} = 0.050 \text{ moles of gases} \]

### Step 5: Calculate the Molar Mass of the Gaseous Mixture
Now, we need to find the average molar mass of the gaseous mixture of `SO2` and `SO3`.

- Molar mass of `SO2` = 32.07 + (2 × 16.00) = 64.07 g/mol
- Molar mass of `SO3` = 32.07 + (3 × 16.00) = 80.07 g/mol

The average molar mass of the gaseous mixture is calculated as follows:
\[ \text{Average Molar Mass} = \frac{\text{Molar mass of } SO2 + \text{Molar mass of } SO3}{\text{Number of moles of gases}} \]
\[ \text{Average Molar Mass} = \frac{64.07 + 80.07}{2} = \frac{144.14}{2} = 72.07 \text{ g/mol} \]

### Final Answer
The average molar mass of the gaseous mixture is approximately **72 g/mol**.

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