Question

Determine the mass of product that will precipitate when \(150.0 \mathrm{~mL} 0.2753 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(220.5 \mathrm{~mL}\) of \(0.1873 \mathrm{M} \mathrm{NaI}(a q)\) are combined.

Short answer

9.524 grams of PbI2 will precipitate.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction between lead(II) nitrate and sodium iodide is represented by the chemical equation: \[ \text{Pb(NO}_3\text{)}_2 (aq) + 2\text{NaI} (aq) \rightarrow \text{PbI}_2 (s) + 2\text{NaNO}_3 (aq) \] Lead(II) iodide (PbI\(_2\)) is the solid that precipitates.

Calculate the Moles of Each Reactant

First, calculate the moles of \(\text{Pb(NO}_3\text{)}_2\) using its volume and molarity: \[ \text{moles of } \text{Pb(NO}_3\text{)}_2 = 0.1500 ext{ L} \times 0.2753 ext{ M} = 0.041295 ext{ moles} \]Next, calculate the moles of \(\text{NaI}\):\[ \text{moles of NaI} = 0.2205 ext{ L} \times 0.1873 ext{ M} = 0.04128865 ext{ moles} \]

Determine the Limiting Reactant

The balanced equation shows that 1 mole of \(\text{Pb(NO}_3\text{)}_2\) reacts with 2 moles of \(\text{NaI}\). Therefore, we need twice as many moles of \(\text{NaI}\) compared to \(\text{Pb(NO}_3\text{)}_2\) for a complete reaction.Given:- Moles of \(\text{Pb(NO}_3\text{)}_2\): 0.041295- Moles of \(\text{NaI}\): 0.04128865 (which is less than twice the moles of \(\text{Pb(NO}_3\text{)}_2\))Thus, \(\text{NaI}\) is the limiting reactant.

Calculate the Moles of Precipitate Formed

Since \(\text{NaI}\) is the limiting reactant, it determines the amount of \(\text{PbI}_2\) that can form. From the stoichiometry of the reaction, 2 moles of \(\text{NaI}\) produce 1 mole of \(\text{PbI}_2\). Thus:\[ \text{moles of PbI}_2 = \frac{0.04128865}{2} = 0.020644325 \text{ moles} \]

Convert Moles of Precipitate to Mass

Calculate the mass of \(\text{PbI}_2\) using its molar mass (\(\text{PbI}_2\): 461.01 g/mol):\[ \text{mass of PbI}_2 = 0.020644325 \text{ moles} \times 461.01 \text{ g/mol} = 9.524 \text{ g} \]

Key concepts

Balanced Chemical Equation

A balanced chemical equation is fundamental in understanding how molecules interact in a chemical reaction. It represents the reactants and products, along with their respective quantities, in the reaction. For instance, in our exercise, the balanced chemical equation is given as:
\[\text{Pb(NO}_3\text{)}_2 (aq) + 2\text{NaI} (aq) \rightarrow \text{PbI}_2 (s) + 2\text{NaNO}_3 (aq)\]This equation tells us that one molecule of lead(II) nitrate reacts with two molecules of sodium iodide to produce one molecule of lead(II) iodide and two molecules of sodium nitrate.

• The coefficients (numbers in front of molecules) indicate the proportionate number of moles involved in the reaction.
• Here, it shows that for every 1 mole of \( \text{Pb(NO}_3\text{)}_2 \) used, 2 moles of \( \text{NaI} \) are required.
• Furthermore, it indicates that \( \text{PbI}_2 \) forms as a precipitate, which is evident by its solid (s) notation.

Knowing how to write and balance chemical equations helps predict the products and quantities formed and is crucial for stoichiometric calculations.

Limiting Reactant

The limiting reactant is the substance that gets completely consumed first in a chemical reaction, thus limiting the amount of product formed. Determining the limiting reactant is essential because it dictates the maximum yield of the reaction.
In our given reaction, which involves \( \text{Pb(NO}_3\text{)}_2 \) and \( \text{NaI} \), the stoichiometry tells us that 1 mole of \( \text{Pb(NO}_3\text{)}_2 \) requires 2 moles of \( \text{NaI} \). By comparing the available moles:

• Moles of \( \text{Pb(NO}_3\text{)}_2 \): 0.041295
• Moles of \( \text{NaI} \): 0.041289 (less than what's needed for the available \( \text{Pb(NO}_3\text{)}_2 \))

From this calculation, we see that \( \text{NaI} \) is the limiting reactant. It will entirely react with \( \text{Pb(NO}_3\text{)}_2 \), thereby determining the amount of \( \text{PbI}_2 \) that can be formed. Understanding this concept can help maximize product formation in industrial and laboratory settings.

Moles Calculation

Calculating moles is a central part of stoichiometric calculations in chemistry. Moles provide a way to quantify chemical substances in a reaction, allowing for an easy comparison and balance of equations. To find moles:
You use the formula:

• \( \text{moles} = \text{volume in liters} \times \text{molarity} \)

For \( \text{Pb(NO}_3\text{)}_2 \):
\[\text{moles of Pb(NO}_3\text{)}_2 = 0.1500 \text{ L} \times 0.2753 \text{ M} = 0.041295 \text{ moles}\]And for \( \text{NaI} \):
\[\text{moles of NaI} = 0.2205 \text{ L} \times 0.1873 \text{ M} = 0.04128865 \text{ moles}\] Moles calculation is completed to determine how much of each reactant will combine and to work out the limiting reactant, which affects the output of the product. It aids understanding on both a molecular/micro level and a practical, experimental level.

Precipitate Formation

Precipitate formation in chemical reactions occurs when an insoluble solid emerges from a solution. This happens when certain ions in a solution come together to form an insoluble compound. In this exercise, when \( \text{Pb(NO}_3\text{)}_2 \) reacts with \( \text{NaI} \), lead(II) iodide (\( \text{PbI}_2 \)) forms as a precipitate, signifying a chemical change.
Key considerations:

• Precipitates are denoted by the solid (\( s \)) label in a chemical equation, indicating their lack of solubility in the reaction mixture.
• The reality of a precipitate forming is visually noticeable as a cloudy or solid formation in the solution.
• Using stoichiometry, we calculate moles of \( \text{PbI}_2 \) formed based on the limiting reactant \( \text{NaI} \).

The calculation can be illustrated as:
\[\text{moles of PbI}_2 = \frac{0.04128865}{2} = 0.020644325 \text{ moles}\]Finally, converting moles to mass using molar mass gives us the real, tangible mass of the precipitate. Understanding this concept allows chemists to predict the extent and limitation of reaction results.