When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead iodide are formed. The balanced chemical equation is: Pb(NO3)2 (aq) + 2 NaI (aq) -> PbI2 (s) + 2 NaNO3 (aq) If we start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, we can calculate how many grams of sodium nitrate can be formed. To find the limiting reagent, we need to compare the number of moles of each reactant. The molar mass of Pb(NO3)2 is 331.2 g/mol, and the molar mass of NaI is 149.9 g/mol. For lead (II) nitrate: 25.0 g Pb(NO3)2 * (1 mol Pb(NO3)2 / 331.2 g Pb(NO3)2) = 0.0755 mol Pb(NO3)2 For sodium iodide: 15.0 g NaI * (1 mol NaI / 149.9 g NaI) = 0.1001 mol NaI According to the balanced equation, the ratio of Pb(NO3)2 to NaI is 1:2. This means that for every 1 mol of Pb(NO3)2, we need 2 mol of NaI. Therefore, the limiting reagent is Pb(NO3)2 because we have less moles of it compared to NaI. To find the grams of sodium nitrate formed, we can use the stoichiometry of the balanced equation. The molar mass of NaNO3 is 85.0 g/mol. For sodium nitrate: 0.0755 mol Pb(NO3)2 * (2 mol NaNO3 / 1 mol Pb(NO3)2) * (85.0 g NaNO3 / 1 mol NaNO3) = 12.9 g NaNO3 Therefore, 12.9 grams of sodium nitrate can be formed. To find the grams of lead (II) iodide formed, we can use the stoichiometry of the balanced equation. The molar mass of PbI2 is 461.0 g/mol. For lead (II) iodide: 0.0755 mol Pb(NO3)2 * (1 mol PbI2 / 1 mol Pb(NO3)2) * (461.0 g PbI2 / 1 mol PbI2) = 34.8 g PbI2 Therefore, 34.8 grams of lead (II) iodide is formed. If 6 grams of sodium nitrate are actually formed in the reaction described in problem #2, we can calculate the percent yield of this reaction. The theoretical yield of sodium nitrate is 12.9 grams (as calculated earlier). The actual yield is given as 6 grams. Percent yield = (actual yield / theoretical yield) * 100 Percent yield = (6 g / 12.9 g) * 100 = 46.5% Therefore, the percent yield of this reaction is 46.5%.
The Correct Answer and Explanation is:
ChatGPT said:
To determine the percent yield of sodium nitrate in the reaction between lead (II) nitrate and sodium iodide, let’s break down the steps and calculations:
Step 1: Balanced Equation
The balanced chemical equation for the reaction is:
Pb(NO3)2(aq)+2NaI(aq)→PbI2(s)+2NaNO3(aq)\text{Pb(NO}_3)_2 (aq) + 2 \text{NaI} (aq) \rightarrow \text{PbI}_2 (s) + 2 \text{NaNO}_3 (aq)Pb(NO3)2(aq)+2NaI(aq)→PbI2(s)+2NaNO3(aq)
Step 2: Moles of Reactants
We start by calculating the number of moles of each reactant:
- For lead (II) nitrate (Pb(NO₃)₂):
- Molar mass of Pb(NO₃)₂ = 331.2 g/mol
- Moles of Pb(NO₃)₂ = 25.0 g331.2 g/mol=0.0755 mol\frac{25.0 \text{ g}}{331.2 \text{ g/mol}} = 0.0755 \text{ mol}331.2 g/mol25.0 g=0.0755 mol
- For sodium iodide (NaI):
- Molar mass of NaI = 149.9 g/mol
- Moles of NaI = 15.0 g149.9 g/mol=0.1001 mol\frac{15.0 \text{ g}}{149.9 \text{ g/mol}} = 0.1001 \text{ mol}149.9 g/mol15.0 g=0.1001 mol
Step 3: Limiting Reagent
From the balanced equation, the molar ratio of Pb(NO₃)₂ to NaI is 1:2. This means:
- For every 1 mole of Pb(NO₃)₂, we need 2 moles of NaI.
- The number of moles of NaI required to fully react with 0.0755 mol of Pb(NO₃)₂ is:
0.0755 mol Pb(NO₃)₂×2=0.151 mol NaI0.0755 \text{ mol Pb(NO₃)₂} \times 2 = 0.151 \text{ mol NaI}0.0755 mol Pb(NO₃)₂×2=0.151 mol NaI
Since we only have 0.1001 mol of NaI (less than 0.151 mol required), NaI is in excess, and Pb(NO₃)₂ is the limiting reagent.
Step 4: Theoretical Yield of Sodium Nitrate
According to the balanced equation:
- 1 mole of Pb(NO₃)₂ produces 2 moles of NaNO₃.
- Moles of NaNO₃ produced from 0.0755 mol of Pb(NO₃)₂ = 0.0755 mol Pb(NO₃)₂×2=0.151 mol NaNO₃0.0755 \text{ mol Pb(NO₃)₂} \times 2 = 0.151 \text{ mol NaNO₃}0.0755 mol Pb(NO₃)₂×2=0.151 mol NaNO₃
The molar mass of NaNO₃ is 85.0 g/mol, so the theoretical yield of NaNO₃ is:
- 0.151 mol NaNO₃×85.0 g/mol=12.9 g NaNO₃0.151 \text{ mol NaNO₃} \times 85.0 \text{ g/mol} = 12.9 \text{ g NaNO₃}0.151 mol NaNO₃×85.0 g/mol=12.9 g NaNO₃
Step 5: Actual Yield and Percent Yield
The actual yield of sodium nitrate is given as 6 grams. The percent yield can be calculated using the formula:
Percent Yield=(Actual YieldTheoretical Yield)×100\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100Percent Yield=(Theoretical YieldActual Yield)×100
Substituting the values:
Percent Yield=(6 g12.9 g)×100=46.5%\text{Percent Yield} = \left( \frac{6 \text{ g}}{12.9 \text{ g}} \right) \times 100 = 46.5\%Percent Yield=(12.9 g6 g)×100=46.5%
Conclusion
The percent yield of sodium nitrate in this reaction is 46.5%. This means that only 46.5% of the theoretical amount of sodium nitrate was actually produced, which could be due to experimental losses, impurities, or incomplete reactions.
