Question

What mass of barium sulfate can be produced when \(100.0 \mathrm{mL}\) of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?

Short answer

When 100.0 mL of 0.100 M barium chloride is mixed with 100.0 mL of 0.100 M iron(III) sulfate, approximately 4.67 g of barium sulfate can be produced.

Step-by-step solution

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between barium chloride (BaCl2) and iron(III) sulfate (Fe2(SO4)3) is: \[ BaCl_2(aq) + Fe_2(SO_4)_3(aq) \rightarrow 2BaSO_4(s) + 6FeCl_3(aq) \]

Converting volumes to moles

Given 100.0 mL of 0.100 M barium chloride and 100.0 mL of 0.100 M iron(III) sulfate, we can convert the volume to moles using the formula: Moles = Molarity × Volume (in liters) For barium chloride: Moles of BaCl2 = (0.100 mol/L) × (100.0 mL / 1000 mL/L) = 0.0100 mol For iron(III) sulfate: Moles of Fe2(SO4)3 = (0.100 mol/L) × (100.0 mL / 1000 mL/L) = 0.0100 mol

Finding the limiting reactant

In the balanced equation, we can see that 1 mole of BaCl2 reacts with 0.5 moles of Fe2(SO4)3: 1 mol BaCl2 : 0.5 mol Fe2(SO4)3 Divide the number of moles of each reactant by their respective stoichiometric coefficients: For BaCl2: 0.0100 mol / 1 = 0.0100 For Fe2(SO4)3: 0.0100 mol / 0.5 = 0.0200 Since 0.0100 < 0.0200, BaCl2 is the limiting reactant as it has the lowest ratio.

Calculate the moles of product (barium sulfate) formed

According to the balanced equation, 1 mole of BaCl2 produces 2 moles of barium sulfate: 1 mol BaCl2 --> 2 mol BaSO4 Using the limiting reactant (BaCl2), we can find the moles of barium sulfate produced: Moles of BaSO4 = (moles of BaCl2) × (2 moles BaSO4 / 1 mol BaCl2) = (0.0100 mol) × 2 = 0.0200 mol

Calculate the mass of barium sulfate formed

Using the molar mass of barium sulfate (BaSO4), we can convert moles to mass: Molar mass of BaSO4 = 233.43 g/mol Mass of BaSO4 = Moles of BaSO4 × Molar mass of BaSO4 = (0.0200 mol) × (233.43 g/mol) = 4.6686 g Therefore, the mass of barium sulfate that can be produced when 100.0 mL of 0.100 M barium chloride is mixed with 100.0 mL of 0.100 M iron(III) sulfate is approximately 4.67 g.

Key concepts

Chemical Reactions

Chemical reactions are the processes in which substances, called reactants, transform into products. This transformation involves the breaking and forming of chemical bonds, leading to the production of new substances with different properties.
In the context of Barium Sulfate Precipitation, a chemical reaction occurs when solutions of barium chloride and iron(III) sulfate are mixed.
The substances involved rearrange their atoms to form new products, which in this case are barium sulfate (a solid precipitate) and iron(III) chloride, remaining in solution.
The balanced chemical equation for this reaction is:

• BaCl_2(aq) + Fe_2(SO_4)_3(aq) → 2BaSO_4(s) + 6FeCl_3(aq)

This equation indicates that one molecule of barium chloride reacts with one-half molecule of iron(III) sulfate to produce two molecules of barium sulfate and six molecules of iron(III) chloride.

Limiting Reactant

The concept of the limiting reactant is key to understanding chemical reactions thoroughly.
The limiting reactant is the substance that will be consumed first in a chemical reaction, thereby determining how much product can be formed. It "limits" the reaction.
In our problem, after calculating the moles of barium chloride and iron(III) sulfate, we identify which one is the limiting reactant by comparing moles available versus the stoichiometric ratio required for the reaction.

• For this reaction: 1 mole of BaCl2 reacts with 0.5 moles of Fe2(SO4)3.
• We have 0.0100 mol of each in this case, but we need 0.0200 mol of Fe2(SO4)3 for 0.0100 mol of BaCl2.
• Hence, BaCl2 is the limiting reactant because it has the lowest calculated ratio.

Recognizing the limiting reactant is crucial because it helps predict how much product will be formed, as the reaction cannot proceed without it.

Stoichiometry

Stoichiometry is a branch of chemistry that focuses on the quantitative relationships between the substances involved in chemical reactions.
It allows us to predict how much of each substance is required or produced in a given reaction.
In this exercise, stoichiometry helps us understand how barium chloride and iron(III) sulfate react in specific proportions to deliver a certain amount of barium sulfate. Firstly, we determine the moles of each reactant from their concentrations and volumes (Molarity × Volume).
Then, using the balanced equation:

• 1 mol BaCl2 reacts with 0.5 mol Fe2(SO4)3.
• 0.0100 mol BaCl2 results in 0.0200 mol BaSO4.

Here, stoichiometry helps in identifying that 2 moles of barium sulfate are produced per mole of BaCl2. It highlights the importance of having balanced chemical equations for predicting product yields.

Solution Chemistry

Solution chemistry is essential for understanding reactions that occur in aqueous media, where substances are dissolved in water.
It deals with concepts such as concentration, molarity, and the behavior of dissolved ions.
In our scenario, both barium chloride and iron(III) sulfate are initially in solution form, allowing their ions to be free to move and react. By understanding solution chemistry, we can calculate the concentration of solutions in terms of molarity (moles of solute per liter of solution) and map out how reactions proceed as ions interact.
When barium chloride is mixed with iron(III) sulfate, their ions precipitate to form insoluble barium sulfate:

• The aqueous nature allows ions to form new products easily.
• Insoluble barium sulfate precipitates out, visible as solid particles.
• Iron(III) chloride remains dissolved in the aqueous solution.

Thus, solution chemistry helps to explain the behavior, interaction, and resultant precipitation of ions from dissolved chemicals in water.