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 \(100.0\) \(\mathrm{mL}\) of a \(0.100 \mathrm{M}\) solution of iron(III) sulfate?

Short answer

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, 4.67 g of barium sulfate can be produced.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between barium chloride (BaCl2) and iron(III) sulfate (Fe2(SO4)3) is: \[ 3BaCl_2(aq) + 2Fe_2(SO_4)_3(aq) \rightarrow 6BaSO_4(s) + 4FeCl_3(aq) \] In this reaction, three moles of barium chloride react with two moles of iron(III) sulfate to form six moles of barium sulfate and four moles of iron(III) chloride.

Calculate the moles of each reactant

We have 100.0 mL of a 0.100 M solution of barium chloride, and 100.0 mL of a 0.100 M solution of iron(III) sulfate. To calculate the moles of each reactant, use the formula: Moles = Molarity × Volume (in L) For barium chloride: Moles = 0.100 mol/L × 0.100 L = 0.0100 mol For iron(III) sulfate: Moles = 0.100 mol/L × 0.100 L = 0.0100 mol

Determine the limiting reactant

To determine the limiting reactant, we'll compare the moles of each reactant to their stoichiometric ratio in the balanced chemical equation. For barium chloride (3 moles of BaCl2 needed for each 2 moles of Fe2(SO4)3): \( \frac{0.0100~mol}{3} = 0.00333 \) For iron(III) sulfate: \( \frac{0.0100~mol}{2} = 0.00500 \) Since the value for barium chloride is smaller, barium chloride is the limiting reactant.

Calculate the moles of the product

In the balanced chemical equation, three moles of barium chloride react to produce six moles of barium sulfate (BaSO4). So, we can use the stoichiometric ratio to find the moles of barium sulfate produced: Moles of barium sulfate = \( \frac{6}{3} \times 0.0100~mol \) = 0.0200 mol

Convert moles of the product to mass

Finally, to convert the moles of barium sulfate to mass, use the formula: Mass = Moles × Molar mass The molar mass of barium sulfate (BaSO4) is: \[ 137.33 (Ba) + 32.07 (S) + 15.999 (O) \times 4 = 233.43~g/mol\] So, the mass of barium sulfate produced is: Mass = 0.0200 mol × 233.43 g/mol = 4.67 g Therefore, 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, 4.67 g of barium sulfate can be produced.

Key concepts

Stoichiometry

Stoichiometry is a crucial aspect of chemistry that allows us to predict the outcomes of chemical reactions. It's like a recipe in cooking, where it dictates the amounts of ingredients needed to get the desired product. When dealing with chemical reactions, stoichiometry helps determine how much reactants are needed and how much product will be formed. By using molar ratios derived from balanced chemical equations, stoichiometry converts quantities of reactants and products into meaningful figures.

• It involves calculations based on the coefficients from the balanced equation.
• It provides a direct relation between mass, moles, and molecules.
• It's essential for finding limiting reactants and excess reactants.

Moreover, understanding stoichiometry assists in ensuring that experiments are efficient and nothing is wasted. It also plays a vital role in fields like pharmaceuticals and industrial production.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed first. This reactant limits the amount of product that can be formed. Think of it as the ingredient that runs out first when baking a cake; you can't bake more cakes without it.

• The limiting reactant is determined by comparing the mole ratios of the reactants with those required by the balanced chemical equation.
• In our exercise, barium chloride is the limiting reactant, as it runs out before iron(III) sulfate.
• Identifying the limiting reactant is crucial for calculating the maximum amount of product possible.

This concept is critical for cost efficiency and resource management in industrial processes, as it helps predict yield and prevent excessive use of other reactants.

Balanced Chemical Equation

A balanced chemical equation is an expression that shows the reactants and products in a chemical reaction with their respective quantities. It's the backbone of stoichiometry and ensures that matter is conserved during any reaction.

• All chemical equations must be balanced to satisfy the law of conservation of mass, meaning the number of atoms for each element on the reactant side should equal the number on the product side.
• In our example, the equation is: \[3BaCl_2(aq) + 2Fe_2(SO_4)_3(aq) \rightarrow 6BaSO_4(s) + 4FeCl_3(aq)\]
• Balancing involves adjusting coefficients to reflect the true stoichiometric relationship between reactants and products.

By using balanced equations, chemists can perform accurate mole-to-mass conversions and predict the outputs of reactions with precision.

Moles Calculation

Moles calculation is a foundational tool in chemistry for quantifying substances. The mole is a standard scientific unit for measuring large quantities of very small entities like atoms or molecules.

• The number of moles is calculated using the formula: \[\text{Moles} = \text{Molarity} \times \text{Volume (in L)}\]
• This concept helps translate laboratory measurements into actionable data for chemical equations.
• For instance, in our problem, 0.100 mol of barium chloride and 0.100 mol of iron(III) sulfate were calculated by their given molarity and volume, which was essential for further calculations.

Understanding moles calculation allows chemists to interchange between mass, volume, and number of atoms or molecules, streamlining the design and interpretation of experiments.