Question

What is the mass and the identity of the precipitate that forms when \(55.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{BaCl}_{2}\) reacts with \(40.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Short answer

The precipitate is barium carbonate (\(\mathrm{BaCO}_3\)), massing 1.09 g.

Step-by-step solution

Write the balanced equation

The reaction between barium chloride (\(\mathrm{BaCl}_2\)) and sodium carbonate (\(\mathrm{Na}_2\mathrm{CO}_3\)) can be described by the balanced chemical equation: \[\mathrm{BaCl}_2 (aq) + \mathrm{Na}_2\mathrm{CO}_3 (aq) \rightarrow \mathrm{BaCO}_3(s) + 2\mathrm{NaCl} (aq)\] Here, barium carbonate (\(\mathrm{BaCO}_3\)) is the precipitate that forms.

Calculate moles of reactants

Using the volume and molarity of the solutions to find moles: - Moles of \(\mathrm{BaCl}_2\): \(\text{Molarity} \times \text{Volume} = 0.100 \ \mathrm{M} \times 0.0550 \ \mathrm{L} = 0.0055 \ \mathrm{mol}\) - Moles of \(\mathrm{Na}_2\mathrm{CO}_3\): \(0.150 \ \mathrm{M} \times 0.0400 \ \mathrm{L} = 0.0060 \ \mathrm{mol}\)

Identify the limiting reactant

The balanced equation shows a 1:1 ratio between \(\mathrm{BaCl}_2\) and \(\mathrm{Na}_2\mathrm{CO}_3\). Therefore:- We have 0.0055 moles of \(\mathrm{BaCl}_2\)- We have 0.0060 moles of \(\mathrm{Na}_2\mathrm{CO}_3\) \(\mathrm{BaCl}_2\) is the limiting reactant because it has fewer moles.

Calculate moles of precipitate

Since \(\mathrm{BaCl}_2\) is the limiting reactant, it will dictate the amount of \(\mathrm{BaCO}_3\) formed. Based on the stoichiometry (1:1), we will form 0.0055 moles of \(\mathrm{BaCO}_3\).

Convert moles of precipitate to mass

Calculate the mass of \(\mathrm{BaCO}_3\) using its molar mass. 1. Molar mass of \(\mathrm{BaCO}_3\): - Ba: 137.33 g/mol - C: 12.01 g/mol - O: 3 × 16.00 g/mol = 48.00 g/mol - Total: \(137.33 + 12.01 + 48.00 = 197.34 \ \mathrm{g/mol}\)2. Mass of \(\mathrm{BaCO}_3\): - \(0.0055 \ \mathrm{mol} \times 197.34 \ \mathrm{g/mol} = 1.0854 \ \mathrm{g}\)

Conclusion

The identity of the precipitate is barium carbonate (\(\mathrm{BaCO}_3\)) and its mass is approximately 1.09 g.

Key concepts

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid, known as a precipitate. In our case, barium chloride (\( \mathrm{BaCl}_2 \)) and sodium carbonate (\( \mathrm{Na}_2\mathrm{CO}_3 \)) are mixed. During this process,

• The \( \mathrm{Ba}^{2+} \) ions from \( \mathrm{BaCl}_2 \) and the \( \mathrm{CO}_3^{2-} \) ions from \( \mathrm{Na}_2\mathrm{CO}_3 \) react to form barium carbonate, \( \mathrm{BaCO}_3 \).
• \( \mathrm{BaCO}_3 \) then separates from the solution as a solid precipitate because it is insoluble in water.

This reaction simplifies to the equation:\[\mathrm{BaCl}_2 (aq) + \mathrm{Na}_2\mathrm{CO}_3 (aq) \rightarrow \mathrm{BaCO}_3(s) + 2\mathrm{NaCl} (aq)\]Understanding precipitation reactions is crucial, as they tell us how different ions interact in a solution. This is especially significant in fields like chemistry, environmental science, and industrial applications where separation of materials occurs.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, dictating the maximum amount of product formed. In our reaction:

• The moles of \( \mathrm{BaCl}_2 \) were calculated as 0.0055 mol.
• The moles of \( \mathrm{Na}_2\mathrm{CO}_3 \) were computed as 0.0060 mol.

The stoichiometry of the reaction is 1:1, meaning it takes equal moles of \( \mathrm{BaCl}_2 \) and \( \mathrm{Na}_2\mathrm{CO}_3 \) to react completely.
Since there are fewer moles of \( \mathrm{BaCl}_2 \), it is identified as the limiting reactant. It controls the amount of barium carbonate (\( \mathrm{BaCO}_3 \)) that can be formed. Hence, the possibility of more precipitation ends when \( \mathrm{BaCl}_2 \) is fully consumed. Identifying the limiting reactant is essential in various chemical processes to maximize efficiency and minimize waste.

Molar Mass Calculation

Calculating molar mass is an essential skill when dealing with chemical reactions, as it allows conversion from moles to mass - critical for determining the quantity of substances involved. For barium carbonate (\( \mathrm{BaCO}_3 \)), we calculate its molar mass by summing the atomic masses of each constituent element:

• Barium (Ba): Atomic mass is 137.33 g/mol.
• Carbon (C): Atomic mass is 12.01 g/mol.
• Oxygen (O): Since there are three oxygens, it's \( 3 \times 16.00 \) g/mol = 48.00 g/mol.

Thus, the total molar mass of \( \mathrm{BaCO}_3 \) is \( 137.33 + 12.01 + 48.00 = 197.34 \) g/mol.
Converting moles to mass involves multiplying the number of moles by the molar mass. Thus, for \( 0.0055 \) moles of \( \mathrm{BaCO}_3 \):\[0.0055 \ \text{mol} \times 197.34 \ \text{g/mol} = 1.0854 \ \text{g}\]The detailed molar mass calculation crucially determines how much of a substance can be expected upon reaction completion.