### Step-by-Step Solution
1. Balanced Chemical Equation and Type of Reaction
The balanced chemical equation for the reaction between potassium iodide (KI) and lead(II) nitrate (Pb(NO₃)₂) is:
[tex]\[ 2 \text{KI} + \text{Pb(NO₃)₂} \rightarrow 2 \text{KNO₃} + \text{PbI₂} \][/tex]
This is a double displacement reaction, where the two compounds exchange ions to form two new compounds.
2. Given Data
- Mass of KI: [tex]\( 15.0 \, \text{g} \)[/tex]
- Mass of Pb(NO₃)₂: [tex]\( 30.0 \, \text{g} \)[/tex]
- Mass of KNO₃ produced: [tex]\( 20.0 \, \text{g} \)[/tex]
3. Molar Mass Calculations
- Molar mass of KI: [tex]\( 166.0 \, \text{g/mol} \)[/tex]
- Molar mass of Pb(NO₃)₂: [tex]\( 331.2 \, \text{g/mol} \)[/tex]
- Molar mass of KNO₃: [tex]\( 101.1 \, \text{g/mol} \)[/tex]
- Molar mass of PbI₂: [tex]\( 461.0 \, \text{g/mol} \)[/tex]
4. Calculating Moles of Reactants
- Moles of KI:
[tex]\[
\text{moles of KI} = \frac{\text{mass of KI}}{\text{molar mass of KI}} = \frac{15.0 \, \text{g}}{166.0 \, \text{g/mol}} \approx 0.09036 \, \text{mol}
\][/tex]
- Moles of Pb(NO₃)₂:
[tex]\[
\text{moles of Pb(NO₃)₂} = \frac{\text{mass of Pb(NO₃)₂}}{\text{molar mass of Pb(NO₃)₂}} = \frac{30.0 \, \text{g}}{331.2 \, \text{g/mol}} \approx 0.09058 \, \text{mol}
\][/tex]
5. Identifying the Limiting Reagent
According to the balanced chemical equation [tex]\( 2 \text{KI} + \text{Pb(NO₃)₂} \rightarrow 2 \text{KNO₃} + \text{PbI₂} \)[/tex], KI and Pb(NO₃)₂ react in a 2:1 molar ratio. Thus, we compare the moles of each reagent accounting for this ratio:
- Effective moles of KI in reaction: [tex]\(\frac{0.09036}{2} \approx 0.04518 \)[/tex]
- Effective moles of Pb(NO₃)₂ in reaction: [tex]\(0.09058 \)[/tex]
The limiting reagent is the one with the smaller value when adjusted for the reaction ratio. Here, [tex]\(0.04518 \, \text{mol}\)[/tex] (adjusted KI) is the limiting reagent.
6. Calculating the Moles and Mass of PbI₂ Produced
- Moles of PbI₂ produced is equal to the moles of the limiting reagent:
[tex]\[
\text{moles of PbI₂} = 0.04518 \, \text{mol}
\][/tex]
- Mass of PbI₂ produced:
[tex]\[
\text{mass of PbI₂} = \text{moles of PbI₂} \times \text{molar mass of PbI₂} = 0.04518 \, \text{mol} \times 461.0 \, \text{g/mol} \approx 20.83 \, \text{g}
\][/tex]
So, the amount of lead(II) iodide (PbI₂) produced is approximately [tex]\( 20.83 \, \text{g} \)[/tex].
