Question

What mass of silver chloride can be prepared by the reaction of $100.0\mathrm{mL}$ of $0.20\mathrm{M}$ silver nitrate with $100.0\mathrm{mL}$ of $0.15M$ calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Short answer

The mass of silver chloride produced is approximately 2.15 g. The final concentrations of each ion in the solution are: $[Ag⁺]=0.025M$, $[Cl⁻]=0M$, $[NO₃⁻]=0.10M$, and $[Ca²⁺]=0.075M$.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between silver nitrate and calcium chloride. AgNO₃(aq) + CaCl₂(aq) → 2 AgCl(s) + Ca(NO₃)₂(aq)

Determine the number of moles of Ag⁺ and Cl⁻ ions initially present

Next, we will calculate the number of moles of silver ions (Ag⁺) and chloride ions (Cl⁻) in the initial solutions. For Ag⁺ ions: Molarity (M) = moles (n) / volume (L) 0.20 M = n / 0.100 L n(AgNO₃) = 0.020 mol For Cl⁻ ions: 0.15 M = n / 0.100 L n(CaCl₂) = 0.015 mol Since there is a 1:2 molar ratio between AgNO₃ and CaCl₂ in the balanced equation, we need to multiply the number of moles of CaCl₂ by 2 to obtain moles of Cl⁻ ions. n(Cl⁻) = 0.015 mol × 2 = 0.030 mol

Determine the limiting reactant

Now, we will find the limiting reactant by comparing the stoichiometric ratio of reactants. Ag⁺ to Cl⁻ ratio = 1:2 n(Ag⁺) : n(Cl⁻) = 0.020 mol : 0.030 mol When we compare the stoichiometric ratio to the actual ratio of moles in the initial solutions, we find that the Ag⁺ ions are present in excess and the Cl⁻ ions are the limiting reactant.

Calculate the mass of silver chloride produced

Using the limiting reactant (Cl⁻ ions), we can calculate the mass of silver chloride formed: From the balanced equation, 2 mol of Cl⁻ ions will produce 1 mol of AgCl. Moles of AgCl produced = n(Cl⁻) / 2 = 0.030 mol / 2 = 0.015 mol Mass of AgCl = moles of AgCl × molar mass of AgCl = 0.015 mol × (107.87 g/mol + 35.45 g/mol) = 0.015 mol × 143.32 g/mol ≈ 2.15 g

Calculate the concentrations of each ion remaining in the solution

Finally, we will determine the concentrations of Ag⁺, NO₃⁻, Ca²⁺, and Cl⁻ ions remaining in the solution after precipitation. Ag⁺ : Moles remaining = n(Ag⁺) - 0.015 mol = 0.005 mol Since the final volume of the solution will be 200 mL (0.200 L): Concentration of Ag⁺ = 0.005 mol / 0.200 L = 0.025 M Cl⁻ : Since all Cl⁻ ions were used as the limiting reactant, the concentration of Cl⁻ ions remaining is 0 M. NO₃⁻ : Moles of NO₃⁻ ions = moles of AgNO₃ initially = 0.020 mol Concentration of NO₃⁻ = 0.020 mol / 0.200 L = 0.10 M Ca²⁺ : Moles of Ca²⁺ ions = moles of CaCl₂ initially = 0.015 mol Concentration of Ca²⁺ = 0.015 mol / 0.200 L = 0.075 M To summarize, the mass of silver chloride produced is approximately 2.15 g, and the final concentrations of each ion in the solution are: [Ag⁺] = 0.025 M, [Cl⁻] = 0 M, [NO₃⁻] = 0.10 M, and [Ca²⁺] = 0.075 M.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that gets completely consumed first, stopping the reaction from continuing because there is no more of that reactant to react with the other substances. The limiting reactant decides the maximum amount of product that can be formed. Determining the limiting reactant is crucial in stoichiometry because it allows us to calculate how much of the product can actually be made.

In the given exercise, we compared the moles of silver ions ( Ag^+ ) and chloride ions ( Cl^- ) present in the initial solutions to find the limiting reactant. We found there were more than enough Ag^+ ions to react with all Cl^- ions available, indicating Cl^- ions are the limiting reactant. This means the reaction will halt when all Cl^- ions have been used up, thus limiting how much silver chloride ( AgCl ) can form.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They provide essential details such as which reactants are used, which products are formed, and their respective quantities.

• In chemical equations, reactants are shown to the left of an arrow, and products are displayed to the right.
• An equation is balanced when the number of atoms for each element is the same on both sides of the equation.

For the reaction in the exercise, the balanced chemical equation is: AgNO₃(aq) + CaCl₂(aq) → 2 AgCl(s) + Ca(NO₃)₂(aq).

This indicates that one mole of silver nitrate reacts with one mole of calcium chloride to produce two moles of silver chloride and one mole of calcium nitrate. Balancing chemical equations ensures the law of conservation of mass is satisfied, meaning no atoms are lost or gained during the reaction.

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. Calculating molarity is essential for determining how much reactant is available in a reaction, as we saw in the exercise.To find molarity, use the formula:$$M=\frac{n}{V}$$where M is the molarity, n is the number of moles of solute, and V is the volume of the solution in liters.

In our problem, we calculated the molarity of silver nitrate and calcium chloride, which were given as 0.20 M and 0.15 M, respectively. Using these values, we found the moles of each reactant, which helped us identify the limiting reactant and determine the product amount.

Precipitation Reactions

Precipitation reactions occur when two soluble substances in aqueous solutions react to form an insoluble solid, or precipitate, which settles out of the solution. These types of reactions are crucial in separating different chemicals out of a solution, making them highly useful in various chemical processes. In the exercise, silver chloride ( AgCl ) is the precipitate formed from the reaction between silver nitrate ( AgNO₃ ) and calcium chloride ( CaCl₂ ). It is notable because AgCl is insoluble in water and thus forms a solid out of the otherwise clear solution.

• Precipitation allows separation and purification of compounds.
• The solid precipitate can often be collected by filtration.

Recognizing precipitation reactions is essential for determining the products and understanding the changes occurring in a reaction.