When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed. The balanced chemical equation is: Pb(NO3)2 (aq) + 2 NaI (aq) – PbI2 (s) + 2 NaNO3 (aq) If, during this reaction, 23.2 grams of lead (II) nitrate is mixed with 16.8 grams of sodium iodide, Calculate the theoretical yield of lead(II) iodide (provide your answer to the appropriate number of significant figures).
. As part of the Paracetamol synthesis lab, the students also had to perform the following calculation to demonstrate their knowledge. Are you able to help them work this out?
When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed.
The balanced chemical equation is:
Pb(NO3)2 (aq) + 2 NaI (aq) -PbI2 (s) + 2 NaNO3 (aq)
If, during this reaction, 23.2 grams of lead (II) nitrate is mixed with 16.8 grams of sodium iodide,
Calculate the theoretical yield of lead(II) iodide (provide your answer to the appropriate number of significant figures).
The Correct Answer and Explanation is :
The theoretical yield of lead(II) iodide (PbI₂) from the given reaction is approximately 25.83 grams.
Explanation
The balanced chemical equation for the reaction between lead(II) nitrate (Pb(NO₃)₂) and sodium iodide (NaI) is: Pb(NO3)2(aq)+2NaI(aq)→PbI2(s)+2NaNO3(aq)\text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{NaI} (aq) \rightarrow \text{PbI}_2 (s) + 2 \text{NaNO}_3 (aq)
Here’s a step-by-step breakdown of the calculation:
- Molar Mass Calculation:
- Lead(II) nitrate (Pb(NO₃)₂): 207.2+2×(14.01+3×16.00)=331.22 g/mol207.2 + 2 \times (14.01 + 3 \times 16.00) = 331.22 \text{ g/mol}
- Sodium iodide (NaI): 22.99+126.90=149.89 g/mol22.99 + 126.90 = 149.89 \text{ g/mol}
- Lead(II) iodide (PbI₂): 207.2+2×126.90=461.00 g/mol207.2 + 2 \times 126.90 = 461.00 \text{ g/mol}
- Converting Masses to Moles:
- Lead(II) nitrate: 23.2 g331.22 g/mol=0.070 moles \frac{23.2 \text{ g}}{331.22 \text{ g/mol}} = 0.070 \text{ moles}
- Sodium iodide: 16.8 g149.89 g/mol=0.112 moles \frac{16.8 \text{ g}}{149.89 \text{ g/mol}} = 0.112 \text{ moles}
- Stoichiometry Check:
- The reaction requires 2 moles of NaI for every mole of Pb(NO₃)₂.
- Calculated need for NaI based on available Pb(NO₃)₂: 0.070 moles Pb(NO3)2×2=0.140 moles NaI0.070 \text{ moles Pb(NO}_3\text{)}_2 \times 2 = 0.140 \text{ moles NaI}
- Determining the Limiting Reagent:
- Since the available moles of NaI (0.112 moles) is less than the needed moles (0.140 moles), NaI is the limiting reagent.
- Calculating the Theoretical Yield:
- The ratio from the balanced equation for PbI₂ is 1:2 (Pb(NO₃)₂ to NaI). With NaI as the limiting reagent, the moles of PbI₂ produced is half the moles of NaI: 0.112 moles2=0.056 moles PbI2 \frac{0.112 \text{ moles}}{2} = 0.056 \text{ moles PbI}_2.
- The mass of PbI₂ formed: 0.056 moles×461.00 g/mol=25.83 grams0.056 \text{ moles} \times 461.00 \text{ g/mol} = 25.83 \text{ grams}.
This calculation demonstrates the importance of understanding stoichiometry and limiting reagents in predicting the outcomes of chemical reactions. The ability to accurately calculate these yields is crucial for efficient resource use in both academic lab settings and industrial chemical production.
Now, let’s generate an image to represent this reaction visually.
Here is a visual representation of the chemical reaction in a lab setting, showing the formation of lead(II) iodide as a yellow precipitate in a beaker. This image also includes elements like the digital scale displaying the weights of the reactants and the chemical equation on a chalkboard, which ties together the theoretical concepts with practical laboratory work.
