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 \(0.100-M\) solution of iron(III) sulfate?

Short answer

The mass of barium sulfate produced when \(100.0 \mathrm{~mL}\) of a \(0.100-M\) solution of barium chloride is mixed with \(100.0 \mathrm{~mL}\) of a \(0.100-M\) solution of iron(III) sulfate is \(7.00 \: g_{BaSO_{4}}\).

Step-by-step solution

Write the balanced chemical equation for the reaction.

First, we need to write the balanced chemical equation for the reaction between barium chloride and iron(III) sulfate: \[BaCl_{2}(aq) + Fe_{2}(SO_{4})_{3}(aq) \rightarrow 2FeCl_{3}(aq) + 3BaSO_{4}(s)\]

Find the moles of reactants involved in the reaction.

We are given the volume and concentration of each solution, so we can use the formula "moles = volume × concentration" to find the moles of barium chloride and iron(III) sulfate: Moles of \(BaCl_{2}\): \(100.0 \times 10^{-3} L \times 0.100 \frac{moles}{L} = 0.0100 moles_{BaCl_{2}}\) Moles of \(Fe_{2}(SO_{4})_{3}\): \(100.0 \times 10^{-3} L \times 0.100 \frac{moles}{L} = 0.0100 moles_{Fe_{2}(SO_{4})_{3}}\)

Determine the limiting reactant.

To find the limiting reactant, we need to divide the moles of each reactant by their respective stoichiometric coefficients from the balanced chemical equation and compare: \(BaCl_{2}\) ratio: \(\frac{0.0100\:moles}{1} = 0.0100\) \(Fe_{2}(SO_{4})_{3}\) ratio: \(\frac{0.0100\: moles}{1} = 0.0100\) Since both ratios are equal, either reactant can be considered the limiting reactant. We will use barium chloride as the limiting reactant for further calculations.

Find the moles of barium sulfate produced.

Using the stoichiometry of the balanced chemical equation, we can find the moles of barium sulfate produced from the moles of the limiting reactant (barium chloride). Using the stoichiometric coefficients, we can see the ratio between \(BaCl_{2}\) and \(BaSO_{4}\) is 1:3: 0.0100 moles of \(BaCl_{2}\) will produce: \(0.0100\:moles_{BaCl_{2}} \times \frac{3\:moles_{BaSO_{4}}}{1\:mole_{BaCl_{2}}} = 0.0300\:moles_{BaSO_{4}}\)

Calculate the mass of barium sulfate produced.

Finally, we can use the molar mass of barium sulfate to find the mass produced: Molar mass of BaSO_{4} = \(137.33 \: g_{Ba} + 32.07 \: g_{S} + 4(16.00 \: g_{O}) = 233.39 \: g_{BaSO_{4}}/mole\) Mass of BaSO_{4} produced = \(0.0300 \: moles_{BaSO_{4}}\times 233.39 \frac{g_{BaSO_{4}}}{mole_{BaSO_{4}}} = 7.00 \: g_{BaSO_{4}}\) So the mass of barium sulfate produced in the reaction is 7.00 g.

Key concepts

Limiting Reactant Calculation

In chemical reactions, the limiting reactant is the substance that is completely consumed first and thus limits the amount of product that can be formed. To calculate the limiting reactant, you can use a process that involves comparing the mole ratio of each reactant used in the reaction to their respective coefficients in the balanced chemical equation.

When you're given the amounts of multiple reactants, you first convert those amounts to moles. Once you have the moles, divide each amount by the stoichiometric coefficient (from the balanced equation) for that reactant. The reactant with the smallest ratio is the limiting reactant. In cases where the ratios are equal (as in the provided exercise), you can choose either reactant to continue with the subsequent calculations. In our example, barium chloride and iron(III) sulfate have equal moles and stoichiometric coefficients, so barium chloride is chosen as the limiting reactant by convention for further calculations.

Molar Mass Calculation

The molar mass of a chemical compound is the mass of one mole of that substance, and it's typically measured in grams per mole (g/mol). It is calculated by adding together the atomic masses of all the atoms present in the molecule, which can be found on the periodic table.

For example, to calculate the molar mass of barium sulfate (BaSO_4), you add the mass of one barium atom, one sulfur atom, and four oxygen atoms to get the total. Using the atomic masses from the periodic table: Ba (137.33g), S (32.07g), and O (16.00g), you can perform the calculation. Full calculation details are shown in the original exercise, yielding a molar mass of 233.39 g/mol for barium sulfate. Understanding molar mass is crucial as it's used in converting between moles and grams, a fundamental aspect of solving stoichiometry problems.

Chemical Reaction Stoichiometry

Stoichiometry is the part of chemistry that deals with the relative quantities of reactants and products in chemical reactions. For chemical reaction stoichiometry, a balanced chemical equation is essential as it provides the ratio in which reactants combine to form products. These ratios are the stoichiometric coefficients. They tell you how many moles of each reactant are needed to produce a certain number of moles of a product.

In our exercise, the balanced chemical equation shows that 1 mole of barium chloride reacts with 1 mole of iron(III) sulfate to produce 3 moles of barium sulfate and 2 moles of iron(III) chloride. These stoichiometric coefficients are used to find mole-to-mole conversions and guide the rest of the stoichiometry calculations, such as determining the limiting reactant and the theoretical yield.

Mole-to-Mass Conversion

The relationship between moles and mass is a fundamental concept in stoichiometry. To perform a mole-to-mass conversion, you multiply the number of moles by the molar mass of the substance. Conversely, to convert from mass to moles, you would divide the mass by the molar mass.

In the exercise, after determining the number of moles of barium sulfate produced (0.0300 moles) based on the limiting reactant, the next step is to find out the mass of that product. This is done by using the previously calculated molar mass (233.39 g/mol). By multiplying the moles of barium sulfate by its molar mass, the mass of barium sulfate produced can be obtained, which in the provided problem is 7.00 grams. This conversion is critical in stoichiometry to relate quantitative data from the molecular scale to the macroscopic scale.