Question

When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, \(\mathrm{PbCrO}_{4}\), forms. How many grams of lead chromate form when a \(1.00-\mathrm{g}\) sample of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) solution?

Short answer

When a \(1.00-\mathrm{g}\) sample of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) solution, \(0.975 \ g\) of lead chromate forms.

Step-by-step solution

Write the balanced chemical equation

The chemical equation for the reaction between lead(II) nitrate and potassium chromate is: \[Pb(NO_{3})_{2} (aq) + K_{2}CrO_{4} (aq) \rightarrow PbCrO_{4} (s) + 2KNO_{3} (aq)\]

Convert the mass of lead(II) nitrate to moles

To convert the mass of lead(II) nitrate to moles, we should calculate the molar mass of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \). The molar mass of \(\mathrm{Pb(NO_{3})_{2}}\) is \(331.2 \mathrm{~g/mol}\). Given mass = \(1.00 \mathrm{~g}\) \[Moles \ of \ Pb(NO_{3})_{2} = \frac{1.00 \ g}{331.2 \ g/mol} = 0.00302\ mol \]

Convert the volume of potassium chromate solution to moles

We are given that the solution has a molarity of \(1.00 \ M\) and the volume of the solution is \(25.0 \ mL\) (which is equal to \(0.025 \ L\)). \[Moles \ of \ K_{2}CrO_{4} = Molarity \ × \ Volume = 1.00M \ × \ 0.025L = 0.025\ mol \]

Determine the limiting reactant

To determine the limiting reactant, we should compare the mole ratio of the two reactants. According to the balanced chemical equation: \[1\ mol\ of\ Pb(NO_{3})_{2} \ reacts\ with\ 1\ mol\ of\ K_{2}CrO_{4}\] In our case: \(0.00302 \ mol \ of \ Pb(NO_{3})_{2}\) and \(0.025 \ mol \ of \ K_{2}CrO_{4}\) Since we have fewer moles of \(Pb(NO_{3})_{2}\) than \(K_{2}CrO_{4}\), the limiting reactant is \(Pb(NO_{3})_{2}\).

Calculate the mass of lead chromate formed

According to the balanced chemical equation, \(1 \ mol \ of \ Pb(NO_{3})_{2}\) reacts with \(1 \ mol \ of \ K_{2}CrO_{4}\) to form \(1 \ mol \ of \ PbCrO_{4}\). Since we have \(0.00302 \ mol \ of \ Pb(NO_{3})_{2}\) (the limiting reactant) we can calculate the moles of \(PbCrO_{4}\) formed as follows: Moles of \(PbCrO_{4}\) = Moles of \(Pb(NO_{3})_{2}\) = \(0.00302 \ mol\) Now, we can calculate the mass of lead chromate formed. The molar mass of \(PbCrO_{4}\) is \(323.2 \mathrm{~g/mol}\). Mass of \(PbCrO_{4}\) = Moles of \(PbCrO_{4}\) × Molar mass of \(PbCrO_{4}\) \[=0.00302\ mol \ × \ 323.2 \frac{g}{ mol} = 0.975 \ g\] Therefore, when a \(1.00-\mathrm{g}\) sample of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) solution, \(0.975 \ g\) of lead chromate forms.

Key concepts

Chemical Reactions

Understanding chemical reactions is essential for grasping the principles of stoichiometry. At its core, a chemical reaction is a process that involves the rearrangement of atoms and molecules to create new substances. During this process, reactants are transformed into products. The reaction explored in the exercise showcases a double replacement reaction where lead(II) nitrate and potassium chromate exchange components, resulting in lead(II) chromate and potassium nitrate.

It's crucial to write a balanced chemical equation to respect the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. The balanced equation ensures that the number of atoms for each element is the same on both the reactants and products side. This balance lays the groundwork for further stoichiometric calculations, as evident in the provided exercise.

Limiting Reactant

The concept of the limiting reactant is akin to the bottleneck in a production process; it's the reactant that runs out first and thus determines the amount of product that can be formed. Identifying the limiting reactant is a key step in stoichiometric calculations because it directly influences how much product can be generated from given amounts of reactants. In the exercise, we compared the mole ratios of lead(II) nitrate and potassium chromate to ascertain which reactant would limit the formation of lead chromate.

The process involves a few key steps:

• Calculate the number of moles of each reactant.
• Use the balanced chemical equation to see the required mole ratios for reaction.
• Identify the reactant that has fewer moles than required by the mole ratio; this is the limiting reactant.

By determining that lead(II) nitrate was the limiting reactant, we could then calculate the maximum amount of lead chromate that could form.

Mole Concept

The mole concept is a fundamental cornerstone of stoichiometry and bridges the gap between the microscopic world of atoms and the macroscopic world we observe. One mole is defined as Avogadro's number (\(6.022 \times 10^{23}\) entities) of particles, be they atoms, ions, or molecules. It's a convenient way to count particles just like a dozen refers to twelve items. In chemistry, the number of moles provides a way to translate mass measurements into a count of molecules or atoms, which in turn enables chemists to use the balanced equations to predict reaction yields.

In practical terms, we use the molar mass of a substance, which is the mass of one mole of that substance, to convert between mass and moles. As demonstrated in the original exercise, we first calculate the molar mass of a compound (in this case, lead(II) nitrate), and then use it to find out how many moles a given mass represents. This pivotal step is necessary to perform stoichiometric calculations and to ultimately find out how much product can be produced from a specific amount of reactant.