To solve the problem, we need to determine the value of \( x \) in the compound \( \text{FeCl}_x \) based on the given data. Here are the steps to arrive at the solution:
### Step 1: Calculate the moles of AgCl formed
The mass of AgCl precipitate is given as \( 304.8 \, \text{mg} \). First, we convert this mass into grams:
\[
304.8 \, \text{mg} = 304.8 \times 10^{-3} \, \text{g} = 0.3048 \, \text{g}
\]
Next, we need to find the number of moles of AgCl. The molar mass of AgCl is approximately \( 143.32 \, \text{g/mol} \).
Using the formula for moles:
\[
\text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molar mass of AgCl}} = \frac{0.3048 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.00212 \, \text{mol}
\]
### Step 2: Relate moles of AgCl to moles of chloride ions
Since each mole of AgCl corresponds to one mole of chloride ions (\( \text{Cl}^- \)), the moles of chloride ions will also be:
\[
\text{Moles of Cl}^- = 0.00212 \, \text{mol}
\]
### Step 3: Calculate the mass of chloride ions
The mass of chloride ions can be calculated using the molar mass of chlorine, which is approximately \( 35.5 \, \text{g/mol} \):
\[
\text{Mass of Cl}^- = \text{moles of Cl}^- \times \text{molar mass of Cl} = 0.00212 \, \text{mol} \times 35.5 \, \text{g/mol} \approx 0.0753 \, \text{g} = 75.3 \, \text{mg}
\]
### Step 4: Set up the equation for the compound
The total mass of the compound \( \text{FeCl}_x \) is given as \( 134.8 \, \text{mg} \). The mass of the compound can be expressed in terms of \( x \):
\[
\text{Mass of FeCl}_x = \text{mass of Fe} + \text{mass of Cl}
\]
The molar mass of the compound \( \text{FeCl}_x \) can be expressed as:
\[
\text{Molar mass of FeCl}_x = 56 \, \text{g/mol} + 35.5 \, \text{g/mol} \times x
\]
### Step 5: Relate the mass of the compound to the mass of chloride
We can express the mass of chloride in terms of the mass of the compound:
\[
\text{Mass of Cl} = \frac{\text{mass of compound}}{\text{molar mass of compound}} \times x \times 35.5
\]
Setting this equal to the mass of chloride we calculated:
\[
75.3 \, \text{mg} = \frac{134.8 \, \text{mg}}{56 + 35.5x} \times x \times 35.5
\]
### Step 6: Solve for \( x \)
Now we can rearrange and solve for \( x \):
1. Convert all masses to grams for consistency.
2. Rearranging gives us:
\[
75.3 = \frac{134.8 \times 35.5x}{56 + 35.5x}
\]
3. Cross-multiply and simplify to find \( x \).
After calculations, we find \( x \approx 1.8 \). Since \( x \) must be an integer, we round it to the nearest whole number:
\[
x = 2
\]
### Final Answer
The value of \( x \) in the compound \( \text{FeCl}_x \) is \( 2 \), so the compound is \( \text{FeCl}_2 \).
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