Question

What mass of barium sulfate can be produced when 100.0 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

The mass of barium sulfate produced when 100.0 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 is 2.334 g.

Step-by-step solution

Convert volumes to moles of reactants

We are given the volumes (100.0 mL) and concentrations (0.100 M) of both reactants, so we can calculate the moles of each reactant by using the formula: moles = Molarity × Volume (in Liters) Moles of barium chloride (BaCl_2): moles = 0.100 mol/L × 100.0 mL × (1 L / 1000 mL) = 0.0100 mol Moles of iron(III) sulfate (Fe_2(SO_4)_3): moles = 0.100 mol/L × 100.0 mL × (1 L / 1000 mL) = 0.0100 mol

Identify the limiting reactant

In order to determine the limiting reactant, we must check how many moles of the other reactant are needed for each reactant according to the balanced chemical equation. From the equation: 1 mol BaCl_2 reacts with 1/2 mol Fe_2(SO_4)_3 Moles of Fe_2(SO_4)_3 needed for BaCl_2 = (1/2) × 0.0100 mol = 0.0050 mol Since we have 0.0100 mol of Fe_2(SO_4)_3, there is enough iron(III) sulfate to fully react with barium chloride. Therefore, barium chloride (BaCl_2) is the limiting reactant.

Calculate moles of barium sulfate produced

Now that we know the limiting reactant, we can determine the moles of barium sulfate (BaSO_4) produced using the stoichiometry of the balanced equation. From the equation: 1 mol BaCl_2 reacts with 1 mol of BaSO_4 Moles of BaSO_4 produced = 1 × moles of BaCl_2 = 1 × 0.0100 mol = 0.0100 mol

Calculate mass of barium sulfate

Now that we have the moles of barium sulfate (BaSO_4) produced, we can calculate its mass by using the molar mass: mass = moles × molar mass Molar mass of BaSO_4 = 137.33 (Ba) + 32.07 (S) + 4 × 16.00 (O) = 233.40 g/mol Mass of BaSO_4 produced = 0.0100 mol × 233.40 g/mol = 2.334 g The mass of barium sulfate produced when 100.0 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 is 2.334 g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that determines the amount of product formed. Why is it called "limiting"? Because it limits the extent of the reaction — you can’t make more product than your least available reactant can produce. Everything stops once it’s used up. Understanding which reactant is limiting is key to calculating the correct amount of product.

To figure out the limiting reactant:

• Start with a balanced chemical equation.
• Calculate the moles for each reactant given in the problem.
• Use the stoichiometric ratios from the balanced equation to determine which reactant runs out first.

In the problem at hand, barium chloride is identified as the limiting reactant because the stoichiometric ratio shows it will be used up first, dictating the maximum amount of barium sulfate that can form.

Molarity and Concentration

Molarity is a crucial concept in the context of reactions occurring in solutions. It's a measure of the concentration of a solution. Molarity is defined as the number of moles of solute per liter of solution, typically denoted by the symbol M.

For example, in this exercise, both barium chloride and iron(III) sulfate have a molarity of 0.100 M, indicating that there are 0.100 moles of each compound in one liter of solution. Since we're given 100 mL of each, we must convert this volume to liters for molarity calculations:

First, convert 100 mL to liters (100 mL = 0.1 L). Next, use the formula:

• Moles = Molarity × Volume (in Liters)

Thus, for barium chloride: 0.100 mol/L × 0.1 L = 0.010 moles. This straightforward calculation allows us to understand how much of each reactant is present in the solution.

Mass Calculation in Chemistry

Calculating the mass of a chemical product is a common step after deducing the amount of moles produced, especially when dealing with solid products from reactions in solution. Once we have the moles of a substance, we multiply by its molar mass to find its mass.

The molar mass is derived from the atomic masses of the elements involved. For barium sulfate (BaSO_4), the calculation is as follows:

• Barium (Ba): 137.33 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O, 4 atoms): 4 × 16.00 = 64.00 g/mol

Total molar mass of BaSO_4 = 233.40 g/mol.

Knowing that 0.010 moles of BaSO_4 are produced, we calculate the mass as:
0.010 mol × 233.40 g/mol = 2.334 g. This gives the actual mass of barium sulfate produced in the reaction.

Mole Concept

The mole is a foundational concept in chemistry that helps convey the amount of a substance. It links microscopic scales to macroscopic quantities. Defined as the number of atoms in 12 grams of carbon-12, a mole is equal to Avogadro's number, which is approximately 6.022 × 10^{23}.

Using the mole concept, chemists can easily convert between atoms/molecules and grams, making calculations involving chemical reactions more straightforward. It simplifies the process, allowing us to focus on the stoichiometry of the reaction.

In the context of the exercise, understanding moles helps us determine how much of each reactant is available and how much product will be formed. By knowing that 0.010 moles of barium chloride reacts with 0.010 moles of iron(III) sulfate, we can confidently predict the yield of barium sulfate using stoichiometry.