Problem 27 - Introductory Chemistry Atoms First

Question

How would you prepare \(9.70 \mathrm{~g}\) of \(\mathrm{PbCl}_{2}(\mathrm{~s})\) from a \(0.100 \mathrm{M}\) solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and a \(0.200 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) ?

Step-by-Step Solution

Answer

To prepare 9.70 g of PbCl₂, determine the moles of PbCl₂ needed using its molar mass (0.0350 moles). Next, find the moles of both reactants required for the reaction using stoichiometry (also 0.0350 moles). Then, calculate the required volumes of both solutions: 350 mL of 0.100 M Pb(NO₃)₂ and 175 mL of 0.200 M CaCl₂. Mix the calculated volumes of the solutions in a reaction vessel, and a white precipitate of PbCl₂ will form. Finally, filter, wash, and dry the precipitate to obtain 9.70 g of PbCl₂.

Step 1: Convert the mass of PbCl₂ to moles

Divide the given mass of PbCl₂ by its molar mass to find the moles. The molar mass of PbCl₂ = 207.2 g/mol (Pb) + 2 × 35.45 g/mol (Cl) = 277.1 g/mol Moles of PbCl₂ = \(\frac{9.70 g}{277.1 g/mol}\) = 0.0350 moles

Step 2: Use stoichiometry to find the moles of both reactants required

The balanced equation for the reaction is: Pb(NO₃)₂(aq) + CaCl₂(aq) → 2 KNO₃(aq) + PbCl₂(s) From the balanced equation, 1 mole of Pb(NO₃)₂ reacts with 1 mole of CaCl₂ to produce 1 mole of PbCl₂. Therefore, moles of Pb(NO₃)₂ and moles of CaCl₂ required for the reaction are also 0.0350 moles.

Step 3: Calculate the volumes of the given solutions required

Using the molarity formula, we can find the required volumes of both solutions: Molarity (M) = moles/volume(L) For Pb(NO₃)₂ solution: 0.100 M = 0.0350 moles/volume(L) Volume of Pb(NO₃)₂ solution = 0.0350 moles/0.100 M = 0.350 L or 350 mL For CaCl₂ solution: 0.200 M = 0.0350 moles/volume(L) Volume of CaCl₂ solution = 0.0350 moles/0.200 M = 0.175 L or 175 mL

Step 4: Mix the calculated volumes of the solutions

Prepare a reaction vessel to accommodate at least 0.525 L or 525 mL of solution. Pour 350 mL of 0.100 M Pb(NO₃)₂ solution into the vessel, then slowly add 175 mL of 0.200 M CaCl₂ solution while stirring. A white precipitate of PbCl₂ will form, as the reaction between both solutions occurs. Once the volumes of both solutions have been combined and the reaction is complete, filter the precipitate, wash it with distilled water, and allow it to dry to obtain the desired 9.70 g of PbCl₂.

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It tells us how many moles of a solute are present in one liter of solution. To find molarity, use the formula:
\[Molarity (M) = \frac{moles \ of \ solute}{liters \ of \ solution}\]This concept is crucial when you need to mix solutions with precise concentrations. In our exercise, the molarity of the solutions was used to determine the volumes needed to form the desired compound, PbCl₂.

• For the Pb(NO₃)₂ solution, the molarity was 0.100 M.
• For the CaCl₂ solution, it was 0.200 M.

By rearranging the formula, you can calculate the volume of solution needed if you know the number of moles required. For instance, to acquire 0.0350 moles of Pb(NO₃)₂ using a 0.100 M solution, you divide 0.0350 moles by 0.100 M to find the necessary volume, resulting in 350 mL.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately predicting the outcomes of a reaction. It depicts the reactants and products with their respective quantities according to the law of conservation of mass, where matter is neither created nor destroyed.
For our precipitation reaction, the balanced equation is:
\[\text{Pb(NO₃)₂(aq) + CaCl₂(aq) → 2 KNO₃(aq) + PbCl₂(s)}\]This equation shows that one mole of lead(II) nitrate reacts with one mole of calcium chloride to yield one mole of lead(II) chloride, PbCl₂, as a precipitate and two moles of potassium nitrate as a by-product.

• The reaction confirms that the quantities of reactants required to form a specific amount of product are directly proportional.
• Each element’s atoms are equal on both sides of the equation.

Balancing enables precise calculations when determining how much of each reactant is needed to produce a desired amount of product.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction in which two solutions react to form an insoluble solid, known as a precipitate. This reaction occurs when the product of the reactant ions exceeds the solubility product.
In the exercise, mixing solutions of Pb(NO₃)₂ and CaCl₂ results in the formation of solid PbCl₂, a white precipitate. The precipitate indicates that an exchange between ions in the solutions has taken place, forming a new compound.

• The solubility rules help predict when a precipitate will form.
• PbCl₂ is only sparingly soluble in water, which is why it precipitates out.

To complete this type of reaction, the resulting precipitate is usually filtered out of the solution, which can then be washed and dried to obtain the desired compound.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is calculated by adding together the atomic masses from the periodic table for each element in a compound.
For example, the molar mass of lead(II) chloride (PbCl₂) was calculated as:
- 207.2 g/mol for lead (Pb) - Two chlorine (Cl) atoms, each with a mass of 35.45 g/mol, for a total of 70.9 g/mol - This results in a total molar mass of 277.1 g/mol for PbCl₂.

• Knowing the molar mass allows you to convert between grams and moles, which is essential for stoichiometric calculations.
• In our exercise, the conversion enabled us to determine the number of moles from a given 9.70 g mass of PbCl₂ for use in our calculations.

This foundational concept ensures accurate and meaningful chemical computations.