Question

Recognize Cause and Effect Suppose you have 12.56 \(\mathrm{g}\) of a mixture made up of sodium chloride and barium chloride. Explain how you could use a precipitation reaction to determine how much of each compound the mixture contains.

Short answer

To determine the amount of NaCl and BaCl2 in a 12.56 g mixture, use a precipitation reaction with an excess of Na2SO4, causing BaSO4 to form as a precipitate. The balanced equation is: \(BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)\). After adding the mixture to the Na2SO4 solution, filter and measure the mass of the BaSO4 precipitate. Calculate the moles of BaSO4 then determine the mass of BaCl2 in the mixture using its molar mass. Subtract the mass of BaCl2 from 12.56 g to find the mass of NaCl.

Step-by-step solution

Write down the balanced chemical equation for the precipitation reaction

The balanced equation for the reaction between barium chloride (BaCl2) and sodium sulfate (Na2SO4) forming barium sulfate (BaSO4) and sodium chloride (NaCl) is given by: \[ BaCl_2 (aq) + Na_2SO_4 (aq) \rightarrow BaSO_4(s) + 2NaCl (aq) \]

Set up the reaction by mixing the compounds

Prepare a solution containing an excess of sodium sulfate, ensuring that all the barium chloride in the mixture will react to form a precipitate. Then, add the 12.56 g of the NaCl-BaCl2 mixture to this solution.

Filter out the precipitate and measure its mass

After the reaction is complete, the solid BaSO4 will have precipitated out of the solution. To isolate the precipitate, filter the solution, wash the collected solid, and dry it. Once the BaSO4 is dry, measure its mass using a balance.

Calculate the mass of barium chloride and sodium chloride in the mixture

Let the mass of the BaSO4 formed be denoted by M. Use the molar mass of barium sulfate (233.4 g/mol) to determine the moles of BaSO4 formed: \[ moles \, BaSO4 = \frac{M}{233.4} \] Since the stoichiometry of the balanced reaction is 1:1 for BaCl2 and BaSO4, the moles of BaCl2 in the mixture is equal to the moles of BaSO4 formed. Calculate the mass of BaCl2 using its molar mass (137.3 + 2 * 35.45 = 208.2 g/mol): \[ mass \, BaCl_2 = moles \, BaSO4 \times 208.2 \] Next, determine the mass of NaCl in the mixture by subtracting the mass of BaCl2 from the total mass of the mixture (12.56 g): \[ mass \, NaCl = 12.56 - mass \, BaCl_2 \] Now you have determined the amount of sodium chloride and barium chloride in the mixture using a precipitation reaction.

Key concepts

Identifying Compounds

Identifying compounds in a mixture can seem tricky, but with the right chemical reactions, it becomes quite manageable. In our exercise, we start with a mixture of sodium chloride (NaCl) and barium chloride (BaCl2). These are both ionic compounds, which dissolve in water to produce their corresponding ions. To separate and quantify these compounds, we can use a precipitation reaction. By adding a solution of sodium sulfate (Na2SO4) to the mixture, only barium chloride reacts to form a solid precipitate, barium sulfate (BaSO4).
This formation allows us to identify and calculate the barium chloride content because only it forms a distinct solid product, while the sodium chloride remains dissolved in the solution.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions, allowing us to determine the amount of each substance involved. In this scenario, we rely on stoichiometry to interpret the relationship between barium chloride and sodium sulfate. The balanced chemical equation:
\[ BaCl_2 (aq) + Na_2SO_4 (aq) \rightarrow BaSO_4(s) + 2NaCl (aq) \]
shows that one mole of barium chloride reacts with one mole of sodium sulfate to produce one mole of barium sulfate.

• This 1:1 ratio is crucial as it directly links the moles of barium chloride used to the moles of barium sulfate formed.
• By knowing the moles of precipitate (BaSO4), we can backtrack to find moles of BaCl2 initially present, showing stoichiometry's power in chemical calculations.

Chemical Equations

Writing and interpreting chemical equations is vital in chemical reactions, especially when dealing with mixtures like ours. A balanced chemical equation provides us with a concise way to describe the reaction process, showing reactants, products, and their ratios. For instance, our balanced equation:
\[ BaCl_2 (aq) + Na_2SO_4 (aq) \rightarrow BaSO_4(s) + 2NaCl (aq) \]
tells us what substances are involved and in what proportions. The coefficients (numbers in front of compounds) indicate the mole ratios of reactants and products.

• This information is indispensable for predicting the quantities of substances consumed or produced.
• Understanding these ratios ensures we correctly predict the amount of barium sulfate (BaSO4) formed from barium chloride (BaCl2).

Mass Calculation

Mass calculation is an essential aspect of almost every chemical experiment, especially when determining the composition of mixtures. For our mixture, mass calculation allows us to determine how much of each compound is present by using the precipitated barium sulfate's mass. Once we isolate and weigh the BaSO4, we use its molar mass (233.4 g/mol) to find the moles of BaSO4. Since one mole of BaCl2 forms one mole of BaSO4, the moles of BaSO4 give us the moles of BaCl2, which can then be converted into mass using its molar mass (208.2 g/mol).

• The mass of BaCl2 remains crucial to calculate the remaining NaCl in the mixture.
• By subtracting the mass of BaCl2 from the total mixture mass, we find the mass of NaCl, completing the analysis.

Mass calculation combines stoichiometry with equations to provide a detailed picture of a mixture's composition.