Problem 109 - General Chemistry: Principles and Modern Applications

Question

A 99.8 mL sample of a solution that is \(120 \%\) KI by mass \((d=1.093 \mathrm{g} / \mathrm{mL})\) is added to \(96.7 \mathrm{mL}\) of another solution that is \(14.0 \% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass \((d=1.134 \mathrm{g} / \mathrm{mL})\) How many grams of \(\mathrm{PbI}_{2}\) should form? \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{KI}(\mathrm{aq}) \longrightarrow \mathrm{PbI}_{2}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

The reaction should form approximately 21.19 g of PbI2

Step 1: Conversion of Solution Volumes to Masses

Use the provided densities and volumes to calculate the mass of each solution.\n Mass of KI solution = \(\(99.8 mL\) * \(1.093 g/mL\) = \(109.15 g\)\).\n Mass of \(Pb(NO_3)_2\) solution = \(\(96.7 mL\) * \(1.134 g/mL\) = \(109.67 g\)\).

Step 2: Determination of Mass of Reactants

Utilize the mass percentages to find the mass of each reactant.\n Mass of KI = \(\(120 \%\) * \(109.15 g\) / \(100 \%\) = \(130.98 g\)\).\n Mass of \(Pb(NO_3)_2\) = \(\(14.0 \%\) * \(109.67 g\) / \(100 \%\) = \(15.35 g\)\).

Step 3: Conversion of Mass to Moles

Use the molecular masses to convert the mass of the reactants to moles. For KI, molecular mass is \(166.0 g/mol\), and for \(Pb(NO_3)_2\), molecular mass is \(331.2 g/mol\).\nMoles of KI = \(\(130.98 g\) / \(166.0 g/mol\) = \(0.789 mol\)\) \nMoles of \(Pb(NO_3)_2\) = \(\(15.35 g\) / \(331.2 g/mol\) = \(0.046 mol\) \), Note that the reaction consumes KI and \(Pb(NO_3)_2\) in a 2:1 ratio.

Step 4: Identify Limiting Reactant

The limiting reactant is the reactant that is completely used up in the reaction first. With a 2:1 consumption ratio and given the amount of both KI and \(Pb(NO_3)_2\), it is obvious that \(Pb(NO_3)_2\) is the limiting reactant because its quantity is less than half of the quantity of KI.

Step 5: Calculation of Mass of PbI2 formed

The equation shows that 1 mole of \(Pb(NO_3)_2\) produces 1 mole of PbI2. Hence moles of PbI2 produced = moles of \(Pb(NO_3)_2\), which equals \(0.046 mol\). Using the molecular mass of PbI2, \(460.6 g/mol\), calculate the mass of PbI2 produced. \n Mass of PbI2 = Moles of PbI2 * molar mass of PbI2 = \(0.046 mol\) * \(460.6 g/mol\) = \(21.19 g\) of PbI2

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first. As a result, it limits the amount of product that can be formed. Understanding the limiting reactant is crucial since it determines the maximum yield of a reaction.
In our example, we have a reaction between potassium iodide (KI) and lead(II) nitrate \(Pb(NO_3)_2\):

• 2 moles of KI react with 1 mole of \(Pb(NO_3)_2\).
• We calculated the moles of KI as 0.789 moles and \(Pb(NO_3)_2\) as 0.046 moles.
• The reaction is in a 2:1 ratio, meaning 0.046 moles of \(Pb(NO_3)_2\) would consume 0.092 moles of KI.

Thus, \(Pb(NO_3)_2\) is the limiting reactant, as we have more than twice the required amount of KI. This reactant will determine the maximum amount of lead iodide (PbI2) that can be formed.

Stoichiometry

Stoichiometry involves calculating the amounts of reactants and products in chemical reactions. It uses balanced chemical equations to ensure the conservation of mass and charge.
This is illustrated in our scenario:

• The balanced equation is \(Pb(NO_3)_2(aq) + 2 KI(aq) \rightarrow PbI_2(s) + 2 KNO_3(aq)\).
• This equation tells us that 1 mole of \(Pb(NO_3)_2\) reacts with 2 moles of KI to produce 1 mole of PbI2 and 2 moles of KNO3.

By calculating the moles of each reactant and using this balanced equation, we identify how much of each product can form. Stoichiometry allows us to understand the quantities of substances in a reaction and predict how much product will be created given the reactants available.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in g/mol, and is key to converting between masses and moles. It provides a bridge between the macroscopic world and the atomic scale.
For our compounds:

• The molar mass of KI is \(166.0 \text{ g/mol}\).
• The molar mass of \(Pb(NO_3)_2\) is \(331.2 \text{ g/mol}\).
• The molar mass of PbI2 is \(460.6 \text{ g/mol}\).

By using these molar masses, we convert the calculated masses of reactants into moles. This step is crucial for applying stoichiometry because chemical equations are balanced in terms of moles, not mass.

Precipitation Reactions

Precipitation reactions occur when two soluble salts in aqueous solutions combine to form an insoluble solid, known as the precipitate. These reactions play a significant role in various chemical processes.In our case, when solutions containing KI and \(Pb(NO_3)_2\) are mixed:

• Lead iodide (PbI2) forms as a solid precipitate.
• The chemical equation for this reaction is \(Pb(NO_3)_2(aq) + 2 KI(aq) \rightarrow PbI_2(s) + 2 KNO_3(aq)\).

The precipitate, PbI2, is separated from the solution, which confirms its formation during the reaction. This type of reaction is not only important in quantifying reaction yields but also useful in areas like chemical analysis and synthesis.