Problem 215 - Introductory Chemistry Atoms First

Question

Suppose \(20.0 \mathrm{~g}\) of \(\mathrm{NaBr}(\mathrm{s})\) is added to \(50.0 \mathrm{~mL}\) of a \(2.00 \mathrm{M}\) silver nitrate solution. (a) Write a net ionic equation for the formation of the expected precipitate. (b) What is the theoretical yield of precipitate in grams? (c) Suppose only \(15.0 \mathrm{~g}\) of precipitate is isolated. What is the percent yield for the reaction? (d) What is the molar concentration of the excess reactant in the combined solution?

Step-by-Step Solution

Answer

(a) The net ionic equation for the formation of the expected precipitate is: \[Ag^+ + Br^- \rightarrow AgBr(s)\] (b) The theoretical yield of AgBr is 18.8 grams. (c) The percent yield for the reaction is 79.8%. (d) The molar concentration of NaBr, the excess reactant in the combined solution, is 1.88 M.

Step 1: Write the net ionic equation.

Suppose that sodium bromide (NaBr) reacts with silver nitrate (AgNO3) to form sodium nitrate (NaNO3) and silver bromide (AgBr) precipitate. The total ionic equation will then be: \[Na^+ + Br^- + Ag^+ + NO3^- \rightarrow AgBr(s) + Na^+ + NO3^- \] After canceling out the spectator ions (those ions that do not participate directly in the reaction), we are left with the net ionic equation: \[Ag^+ + Br^- \rightarrow AgBr(s)\]

Step 2: Calculate the theoretical yield of AgBr in grams.

The balanced equation tells us that one mole of Ag^+ reacts with one mole of Br^- to produce one mole of AgBr. First, calculate the number of moles of NaBr and AgNO3. Given that the molecular weight of NaBr is approximately 103 g/mole and for AgNO3 is 170 g/mole, we can find: Moles of NaBr = 20.0g / 103g/mol = 0.194 moles Moles of AgNO3 = 2.00 mol/L * 0.050 L = 0.10 moles Since AgNO3 is the limiting reactant (we have less of it), it defines the maximum amount of AgBr that can be formed. Using the stoichiometric ratios from the balanced equation, the theoretical yield of AgBr in moles is therefore 0.10 moles. As the molecular weight of AgBr is approximately 188 g/mole, we find: Theoretical yield of AgBr = 0.10 moles * 188 g/mole = 18.8g

Step 3: Calculate the percent yield for the reaction.

The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%. The actual yield was given as 15.0 grams. Therefore, we get: Percent yield = (15.0g / 18.8g) * 100% = 79.8%

Step 4: Calculate the molar concentration of the excess reactant in the combined solution.

We have NaBr as the excess reactant. We initially had 0.194 moles of it and used up 0.10 moles in the reaction. Thus, we now have 0.094 moles of NaBr left in the solution of 0.050 liters of AgNO3 solution. Therefore, the molar concentration (Molarity) of NaBr is: Molarity = moles of solute / liters of solution = 0.094 moles / 0.050 liters = 1.88 M

Key Concepts

Net Ionic Equations

A net ionic equation succinctly represents a chemical reaction occurring in solution by illustrating only the ions or molecules directly involved in the chemical change, leaving out the spectator ions. Spectator ions are those which do not participate in the actual precipitation, acid-base, or redox reaction. They are present in the reactants and products but remain unchanged and in the same phase.

To construct a net ionic equation, as illustrated in our textbook example, we first write the complete ionic equation showing all the species in a dissolved state separately, unless they form a precipitate or are insoluble. For instance, when sodium bromide is added to a silver nitrate solution, they dissociate into their respective ions. However, silver (Ag+) and bromide (Br-) ions react to form silver bromide (AgBr), a solid precipitate that does not dissociate into ions.

The next step is to eliminate the spectator ions which are identical on both sides of the complete ionic equation. This stripping down leads to the net ionic equation that conveys the essence of the chemical interaction—silver ion plus bromide ion yield silver bromide precipitate, represented as \(Ag^+ + Br^- \rightarrow AgBr(s)\). This concise equation is a powerful tool for understanding and predicting the outcomes of ionic reactions in chemistry.

Percent Yield

Understanding Percent Yield

Percent yield is a crucial concept in chemistry that measures the efficiency of a reaction. It compares the actual yield obtained from an experiment to the theoretical yield predicted by stoichiometry. The theoretical yield is the maximum amount of product that could be formed from the given amounts of reactants, assuming that the reaction goes to completion and there are no losses in the process.

To calculate the percent yield, one must divide the actual yield by the theoretical yield and multiply by 100. This provides a percentage that reflects the reaction's practical success. In our example, the actual yield of AgBr was 15.0 grams, while the theoretical yield was 18.8 grams. Therefore, the percent yield would be calculated as \( \frac{15.0g}{18.8g} \times 100\% = 79.8\% \). A percent yield higher than 100% is typically not possible and suggests errors in the experiment or calculations, while a percent yield lower than 100% indicates that the reaction did not produce as much product as theoretically possible. Percent yield is a vital indicator for the practical viability and economic aspects of chemical processes in the industry.

Molarity Calculations

The Concept of Molarity

Molarity, often represented as 'M', is the measure of the concentration of a solute in a solution. It expresses how many moles of a solute are present in one liter of a solution. Understanding molarity is key for preparing solutions and performing titrations in chemistry.

To calculate the molarity, we divide the number of moles of the solute by the volume of the solution in liters. In the context of our textbook example, we had to find the molarity of the excess reactant \(NaBr\) after the reaction. With 0.094 moles of NaBr remaining in the new total volume of the solution, which is the original 50.0 mL (or 0.050 liters) of the silver nitrate solution, we use the formula: \(\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.094 \text{ moles}}{0.050 \text{ liters}} = 1.88 \text{ M}\).

Molarity calculations are fundamental in predicting how substances will interact in a solution, controlling the stoichiometry of reactants in a reaction, and even in basic laboratory tasks such as diluting concentrations to achieve desired experimental conditions.