Question

An aqueous solution of ammonium sulfide is mixed with an aqueous solution of iron(III) chloride. a. Write a balanced molecular equation, complete ionic equation, and net ionic equation for the two solutions mixed above. Include all phases. b. If \(50.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) ammonium sulfide and \(100.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) iron(III) chloride are mixed, how many grams of precipitate will form? c. What are the concentrations of ammonium ion and iron(III) ion left in solution after the reaction is complete? d. In part b, you started with \(100.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) iron(III) chloride. How would you prepare such a solution in the lab if you started with solid iron(III) chloride?

Short answer

In summary, when ammonium sulfide and iron(III) chloride solutions are mixed, the following equations occur: Balanced molecular equation: \(2(NH_4)_2S(aq) + 6FeCl_3(aq) \rightarrow 12NH_4Cl(aq) + 4Fe_2S_3(s)\) Complete ionic equation: \(4NH_4^{+}(aq) + 2S^{2-}(aq) + 6Fe^{3+}(aq) + 18Cl^-(aq) \rightarrow 12NH_4^{+}(aq) + 12Cl^-(aq) + 4Fe_2S_3(s)\) Net ionic equation: \(2S^{2-}(aq) + 6Fe^{3+}(aq) \rightarrow 4Fe_2S_3(s)\) When \(50.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) ammonium sulfide and \(100.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) iron(III) chloride are mixed, 3.46 grams of iron(III) sulfide precipitate will form. After the reaction, the concentrations of ions left in the solution are: Ammonium ion (\(NH_4^+\)): 0.222 M Iron(III) ion (\(Fe^{3+}\)): 0 M To prepare 100.0 mL of a 0.250 M iron(III) chloride solution, dissolve 4.055 grams of solid iron(III) chloride in 100.0 mL of distilled or deionized water, and mix well in a volumetric flask.

Step-by-step solution

Identify the reaction and predict the products

The reaction is between ammonium sulfide (NH4)2S and iron(III) chloride FeCl3 in aqueous solutions. Both are ionic compounds, so we can predict the formation of new ionic compounds by swapping their respective anions and cations. Therefore, the products will be ammonium chloride (NH4Cl) and iron(III) sulfide (Fe2S3). Ammonium chloride is soluble in water, while iron(III) sulfide is insoluble, which means there will be a precipitate forming in the reaction. Step 2: Write the unbalanced molecular equation.

Write the unbalanced molecular equation

The unbalanced molecular equation representing the reaction is: \((NH_4)_2S(aq) + FeCl_3(aq) \rightarrow NH_4Cl(aq) + Fe_2S_3(s)\) Step 3: Balance the molecular equation.

Balance the molecular equation

The balanced molecular equation is: \(2(NH_4)_2S(aq) + 6FeCl_3(aq) \rightarrow 12NH_4Cl(aq) + 4Fe_2S_3(s)\) Step 4: Write the complete ionic equation.

Write the complete ionic equation

For the complete ionic equation, we separate the soluble ionic compounds into their respective ions: \(4NH_4^{+}(aq) + 2S^{2-}(aq) + 6Fe^{3+}(aq) + 18Cl^-(aq) \rightarrow 12NH_4^{+}(aq) + 12Cl^-(aq) + 4Fe_2S_3(s)\) Step 5: Write the net ionic equation.

Write the net ionic equation

To get the net ionic equation, we cancel out the spectator ions (ions that remain unchanged in the reaction) from the complete ionic equation: \(2S^{2-}(aq) + 6Fe^{3+}(aq) \rightarrow 4Fe_2S_3(s)\) #b. Mass of the precipitate formed from given volumes and concentrations of both solutions# Step 1: Identify the limiting reactant.

Identify the limiting reactant

Given: \(50.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) ammonium sulfide solution \(100.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) iron(III) chloride solution We use the given volume and concentration of the solutions to determine the moles of each reactant: moles of ammonium sulfide = \(50.0 \mathrm{~mL} \times \frac{0.500 \mathrm{mol}}{1000 \mathrm{~mL}} = 0.025 \mathrm{mol}\) moles of iron(III) chloride = \(100.0 \mathrm{~mL} \times \frac{0.250 \mathrm{mol}}{1000 \mathrm{~mL}} = 0.025 \mathrm{mol}\) Next, we compare the mole ratios of the reactants to the balanced equation. The balanced equation states that for every 2 moles of ammonium sulfide, we need 6 moles of iron(III) chloride. Ratio needed = \(\frac{6}{2} = 3\) Ratio given = \(\frac{0.025}{0.025} = 1\) Since the ratio given is less than the needed ratio, iron(III) chloride is the limiting reactant. Step 2: Calculate the mass of the precipitate.

Calculate the mass of the precipitate

Using the limiting reactant iron(III) chloride, we can find out how many moles of precipitate (iron(III) sulfide) will form: \(\frac{4 \mathrm{mol}\ Fe_2S_3}{6 \mathrm{mol}\ FeCl_3} \times 0.025 \mathrm{mol} = 0.01667 \mathrm{mol}\ Fe_2S_3\) The molar mass of iron(III) sulfide (Fe2S3) is: \(2 \times 55.85 (\mathrm{Fe}) + 3 \times 32.07 (\mathrm{S}) = 207.55 \mathrm{g/mol}\) Now we can find the mass of precipitate by multiplying moles of \(Fe_2S_3\) by its molar mass: \(0.01667 \mathrm{mol} \times 207.55 \mathrm{g/mol} = 3.46\ g\) So, 3.46 grams of \(Fe_2S_3\) precipitate will form. #c. Concentrations of remaining ions in the solution after the reaction# Step 1: Calculate remaining moles of ions.

Calculate remaining moles of ions

From the balanced equation, 2 moles of ammonium sulfide react with 6 moles of iron(III) chloride. Hence, 0.025 moles of iron(III) chloride will completely react with \(\frac{0.025}{3} = 0.00833 \mathrm{mol}\) of ammonium sulfide. So, the remaining moles of ammonium sulfide = \(0.025 - 0.00833 = 0.01667 \mathrm{mol}\) Step 2: Calculate the concentration of ions left in solution.

Calculate the concentration of ions left in solution

The total volume of the mixed solution is \(50.0 + 100.0 = 150.0 \mathrm{~mL}\) or \(0.150 \mathrm{L}\) The remaining concentration of ammonium ion (\(NH_4^+\)) is: \(\frac{0.01667 \mathrm{mol} \times 2}{0.150 \mathrm{L}} = 0.222 \mathrm{M}\) The concentration of iron(III) ion (\(Fe^{3+}\)) left is 0M, as all of the Fe ions reacted and formed the precipitate. So the concentrations of the ions left in the solution are: Ammonium ion (\(NH_4^+\)) = 0.222 M Iron(III) ion (\(Fe^{3+}\)) = 0 M #d. Preparing the given iron(III) chloride solution in lab from solid iron(III) chloride# Step 1: Calculate the mass of solid iron(III) chloride required.

Calculate the mass of solid iron(III) chloride required

Our target is to prepare \(100.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) iron(III) chloride solution. To find the mass of the required solid iron(III) chloride, we first find the moles and then use the molar mass of iron(III) chloride (FeCl3). moles of iron(III) chloride = \( 0.100 \mathrm{L} \times 0.250 \mathrm{mol/L} = 0.025 \mathrm{mol} \) The molar mass of \(FeCl_3\) is: \(55.85 (\mathrm{Fe}) + 3 \times 35.45 (\mathrm{Cl}) = 162.20 \mathrm{g/mol}\) Mass required = \(0.025 \mathrm{mol} \times 162.20 \mathrm{g/mol} = 4.055 \mathrm{g}\) Step 2: Prepare the solution.

Prepare the solution

To prepare the solution, carefully weigh out 4.055 grams of solid iron(III) chloride and dissolve it in approximately 100.0 mL of distilled or deionized water. Stir the mixture until the solid has fully dissolved, and then transfer it to a volumetric flask. Fill the flask up to the 100.0 mL mark with distilled or deionized water, stopper the flask, and mix well to ensure a uniform solution.

Key concepts

Stoichiometry

Stoichiometry is the mathematical heart of chemistry. It allows us to predict how one substance in a chemical equation relates to another. In our reaction between ammonium sulfide and iron(III) chloride, stoichiometry helps us determine how much of each reactant is needed to produce a certain amount of product. We begin by writing balanced chemical equations, which show the exact ratio of reactants to products.

Once the equation is balanced, stoichiometry uses these ratios to calculate the moles or grams needed for each reactant, or how much product is formed. It involves conversions between units, like turning molarity and volume into moles, as seen when determining the limiting reactant in our reaction. By calculating moles from the given concentrations and volumes of ammonium sulfide and iron(III) chloride, we can predict the formation of a precipitate, like iron(III) sulfide.

Understanding stoichiometry is key to predicting whether a reaction can occur and to what extent. This is particularly important in practical applications, like preparing solutions in the lab, where precise measurements ensure successful outcomes.

Solution Concentration

Solution concentration indicates how much solute (in this case, ammonium sulfide and iron(III) chloride) is dissolved in a given volume of solvent, usually expressed in molarity (M). Molarity is a crucial concept, as it allows chemists to express the concentration of a solution in terms of moles of solute per liter of solution.

For the reaction under consideration, knowledge of solution concentration is used to calculate the number of moles of each reactant, which is then used to determine how much precipitate is expected. Knowing that you have a 0.500 M ammonium sulfide solution means that there are 0.500 moles of ammonium sulfide in one liter of solution. Similarly, a solution concentration of 0.250 M iron(III) chloride tells us there are 0.250 moles in one liter.

Applying these values, we find the volume needed for the desired amount of reaction by using the formula: \[\text{moles} = \text{molarity} \times \text{volume} \text{ (in L)}\]This calculation determines which reactant is in excess and which is limiting, a concept vital for predicting the outcome of reactions, such as the amount of precipitate formed.

Precipitate Formation

Precipitate formation is a telltale sign of a chemical change where an insoluble solid forms in a solution. In our specific reaction of ammonium sulfide with iron(III) chloride, a precipitate of iron(III) sulfide ( Fe_2S_3) is formed. This occurs because Fe_2S_3 is insoluble in water, unlike many other ionic compounds that remain dissolved.

The ability to predict the formation of a precipitate is essential in various chemical applications. It involves understanding the solubility rules, which tell us which combinations of ions will stay in solution and which will form solids. Iron(III) sulfide's formation results from swapping cations and anions between ammonium sulfide and iron(III) chloride, an action typical in double displacement reactions.

When these ions combine to form iron(III) sulfide, they exceed the solubility product, causing the compound to fall out of solution as a solid. Recognizing and understanding the conditions that cause precipitate formation allows scientists to manipulate and control chemical reactions to achieve desired results in synthesis, purification, and analysis.

Molecular Equations

Molecular equations provide a macroscopic view of the overall reaction without explicitly breaking down each reactant or product into ions. They simplify complex reactions by showcasing all chemicals as complete formulas which helps identify the substances involved in the reaction. In our scenario, the molecular equation for ammonium sulfide mixing with iron(III) chloride helps illustrate this concept.

The balanced molecular equation is:\[2(NH_4)_2S(aq) + 6FeCl_3(aq) \rightarrow 12NH_4Cl(aq) + 4Fe_2S_3(s)\]This equation represents the complete chemical change and is crucial for understanding the stoichiometric relationships between reactants and products. However, it does not show the ionic nature of aqueous compounds or the spectator ions that might be involved.

Molecular equations serve as the first step before writing the ionic equations, providing a foundational framework for further breakdowns into ionic components. This is essential for reactions in aqueous solutions, where ions play a key role in the formation of products and any precipitates. Understanding molecular equations builds the groundwork for further exploration into reaction mechanisms and chemical interactions.