What are the mass and the identity of the precipitate that forms when 55.0 mL of 0.100 M BaCl2 reacts with 40.0 mL of 0.150 M Na2CO3?

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Identify the chemical reaction: \( \text{BaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{BaCO}_3 \downarrow + 2\text{NaCl} \).

Determine the moles of \( \text{BaCl}_2 \) using its concentration and volume: \( \text{moles of } \text{BaCl}_2 = 0.100 \text{ M} \times 0.0550 \text{ L} \).

Determine the moles of \( \text{Na}_2\text{CO}_3 \) using its concentration and volume: \( \text{moles of } \text{Na}_2\text{CO}_3 = 0.150 \text{ M} \times 0.0400 \text{ L} \).

Identify the limiting reactant by comparing the mole ratio from the balanced equation with the calculated moles of each reactant.

Calculate the mass of the precipitate \( \text{BaCO}_3 \) using the moles of the limiting reactant and the molar mass of \( \text{BaCO}_3 \).

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Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble compound, known as a precipitate. In this case, barium chloride (BaCl2) and sodium carbonate (Na2CO3) react to form barium carbonate (BaCO3), which is insoluble in water. Understanding the solubility rules helps predict which products will precipitate.

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Selective Precipitation

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced chemical equations. It allows us to determine the amounts of substances consumed and produced. In this question, stoichiometry is essential for calculating the mass of the precipitate formed from the given concentrations and volumes of the reactants.

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Stoichiometry Concept

Molarity and Volume Calculations

Molarity (M) is a measure of concentration defined as moles of solute per liter of solution. To find the number of moles of each reactant, we multiply the molarity by the volume in liters. This calculation is crucial for determining the limiting reactant and the mass of the precipitate formed in the reaction.

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