`Pb(NO_(3))_(2)` and `KI` reacts in aqueous solution to form an yellow precipitate of `PbI_(2)`. In one series of experiments, the masses of two reactants varied, but the total mass of the two was held constant at `5.0g`. What maximum mass of `PbI_(2)` can be produced in the above experiment :

Explanation

To find the maximum mass of lead(II) iodide (PbI₂) that can be produced from the reaction between lead(II) nitrate (Pb(NO₃)₂) and potassium iodide (KI), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{Pb(NO}_3\text{)}_2 + 2 \text{KI} \rightarrow \text{PbI}_2 + 2 \text{KNO}_3 \]

### Step 2: Determine the molar masses
Next, we need to calculate the molar masses of the reactants and the product:
- Molar mass of PbI₂:
- Pb: 207.2 g/mol
- I: 126.9 g/mol (2 I atoms)
- Total: \( 207.2 + 2 \times 126.9 = 207.2 + 253.8 = 461.0 \, \text{g/mol} \)

- Molar mass of Pb(NO₃)₂:
- Pb: 207.2 g/mol
- N: 14.0 g/mol (2 N atoms)
- O: 16.0 g/mol (6 O atoms)
- Total: \( 207.2 + 2 \times 14.0 + 6 \times 16.0 = 207.2 + 28.0 + 96.0 = 331.2 \, \text{g/mol} \)

- Molar mass of KI:
- K: 39.1 g/mol
- I: 126.9 g/mol
- Total: \( 39.1 + 126.9 = 166.0 \, \text{g/mol} \)

### Step 3: Set up the mass equation
Let \( x \) be the number of moles of Pb(NO₃)₂. According to the balanced equation, 2 moles of KI are required for every mole of Pb(NO₃)₂. The total mass of the reactants is given as 5.0 g:
\[ \text{Mass of Pb(NO}_3\text{)}_2 + \text{Mass of KI} = 5.0 \, \text{g} \]

The mass equation can be expressed as:
\[ 331.2x + 2 \times 166.0x = 5.0 \]
\[ 331.2x + 332.0x = 5.0 \]
\[ 663.2x = 5.0 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{5.0}{663.2} \]
\[ x \approx 0.007539 \, \text{moles of Pb(NO}_3\text{)}_2 \]

### Step 5: Calculate the moles of PbI₂ produced
From the balanced equation, 1 mole of Pb(NO₃)₂ produces 1 mole of PbI₂. Therefore, the moles of PbI₂ produced is also \( 0.007539 \).

### Step 6: Calculate the mass of PbI₂ produced
Now, we can find the mass of PbI₂ produced using its molar mass:
\[ \text{Mass of PbI}_2 = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of PbI}_2 = 0.007539 \, \text{moles} \times 461.0 \, \text{g/mol} \]
\[ \text{Mass of PbI}_2 \approx 3.48 \, \text{g} \]

### Step 7: Round the mass to the nearest integer
Rounding \( 3.48 \) g gives us approximately \( 3 \, \text{g} \).

### Final Answer
The maximum mass of PbI₂ that can be produced is approximately **3 grams**.

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