What is the number of moles of `Fe(OH)_(3)` (s) that can be produced by allowing 1 mole of `Fe_(2)S_(3)`, 2 moles of `H_(2)O` and 3 moles of `O_(2)` to react as : `2Fe_(2)S_(3) + 6H_(2)O + 3O_2 rarr 4Fe(OH)_(3) + 6S` ?

Options

A. 1 mol
B. 1.84 mol
C. 1.34 mol
D. 1.29 mol

Explanation

To solve the problem of how many moles of `Fe(OH)_(3)` can be produced from the reaction of `Fe_(2)S_(3)`, `H_(2)O`, and `O_(2)`, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ 2Fe_{2}S_{3} + 6H_{2}O + 3O_{2} \rightarrow 4Fe(OH)_{3} + 6S \]

### Step 2: Identify the initial moles of reactants
From the problem statement, we have:
- Moles of `Fe_(2)S_(3)`: 1 mole
- Moles of `H_(2)O`: 2 moles
- Moles of `O_(2)`: 3 moles

### Step 3: Determine the stoichiometric coefficients from the balanced equation
From the balanced equation:
- The coefficient for `Fe_(2)S_(3)` is 2
- The coefficient for `H_(2)O` is 6
- The coefficient for `O_(2)` is 3

### Step 4: Calculate the limiting reagent
To find the limiting reagent, we will calculate the number of moles available for each reactant divided by its coefficient:

- For `Fe_(2)S_(3)`:
\[ \text{Available moles} = \frac{1 \text{ mole}}{2} = 0.5 \]

- For `H_(2)O`:
\[ \text{Available moles} = \frac{2 \text{ moles}}{6} = \frac{1}{3} \approx 0.33 \]

- For `O_(2)`:
\[ \text{Available moles} = \frac{3 \text{ moles}}{3} = 1 \]

### Step 5: Identify the limiting reagent
The limiting reagent is the one with the smallest value from the calculations above. Here, the smallest value is `0.33` for `H_(2)O`, which means `H_(2)O` is the limiting reagent.

### Step 6: Calculate the amount of `Fe(OH)_(3)` produced
According to the balanced equation, 6 moles of `H_(2)O` produce 4 moles of `Fe(OH)_(3)`. We can set up a proportion to find out how many moles of `Fe(OH)_(3)` can be produced from 2 moles of `H_(2)O`:

\[ \text{Moles of } Fe(OH)_{3} = \frac{4 \text{ moles } Fe(OH)_{3}}{6 \text{ moles } H_{2}O} \times 2 \text{ moles } H_{2}O \]

Calculating this gives:
\[ \text{Moles of } Fe(OH)_{3} = \frac{4}{6} \times 2 = \frac{8}{6} = 1.33 \text{ moles} \]

### Step 7: Round to appropriate significant figures
Since we are dealing with moles, we can round this to 1.34 moles.

### Final Answer
The number of moles of `Fe(OH)_(3)` produced is **1.34 moles**.

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