Question

A 500.0 -mL sample of \(0.200 \mathrm{M}\) sodium phosphate is mixed with \(400.0 \mathrm{mL}\) of \(0.289 M\) barium chloride. What is the mass of the solid produced?

Short answer

The mass of barium phosphate formed in the reaction between 500.0 mL of 0.200 M sodium phosphate and 400.0 mL of 0.289 M barium chloride is approximately 23.2 g.

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 sodium phosphate and barium chloride. The chemical equation is: \[2 Na_{3}PO_{4} + 3 BaCl_{2} \rightarrow 6 NaCl + Ba_{3}(PO_{4})_{2}\]

Determine the moles of reactants

Now, let's calculate the moles of both reactants. For sodium phosphate: Moles = Molarity × Volume = 0.200 M × 500.0 mL = 0.200 M × 0.500 L = 0.100 moles For barium chloride: Moles = Molarity × Volume = 0.289 M × 400.0 mL = 0.289 M × 0.400 L = 0.1156 moles

Identify the limiting reactant

Using the balanced chemical equation, we can see that for every 2 moles of sodium phosphate, 3 moles of barium chloride are needed. We will use a stoichiometric ratio to determine the limiting reactant. (0.100 moles Na3PO4) / (2 moles Na3PO4) = (0.1156 moles BaCl2) / (3 moles BaCl2) Dividing through, we get: 0.050 / (2/3) = 0.1156 / 3 0.075 = 0.03853 Since 0.03853 < 0.075, barium chloride (BaCl₂) is the limiting reactant.

Calculate the moles of barium phosphate formed

Next, we will use the stoichiometry of the reaction to determine the moles of barium phosphate (Ba₃(PO₄)₂) formed. Using the stoichiometry from the balanced chemical equation: 3 moles BaCl₂ → 1 mole Ba₃(PO₄)₂ So the moles of barium phosphate formed are: (0.1156 moles BaCl₂) × (1 mole Ba₃(PO₄)₂ / 3 moles BaCl₂) = 0.03853 moles Ba₃(PO₄)₂

Calculate the mass of barium phosphate formed

Finally, we will calculate the mass of barium phosphate formed by multiplying the moles of barium phosphate with its molar mass. Molar mass of Ba₃(PO₄)₂ = (3 × Molar mass of Ba) + (2 × (Molar mass of P + (4 × Molar mass of O))) = (3 × 137.33) + (2 × (30.97 + 4 × 16.00)) = 601.98 g/mol Mass of barium phosphate = 0.03853 moles × 601.98 g/mol = 23.2 g (rounded to one decimal place) Thus, the mass of barium phosphate formed in this reaction is 23.2 g.

Key concepts

Chemical Reaction Equations

Understanding chemical reaction equations is essential for grasping the basics of chemical reactions. In our exercise, the starting point is the balanced chemical equation, which is like a recipe for chemists. It shows the reactants on the left, the products on the right, and how they're connected through a reaction.

Each substance is represented by its chemical formula, and the numbers in front of these formulas are called stoichiometric coefficients. They indicate the ratios in which substances react and are formed. For instance, our equation \[2 Na_{3}PO_{4} + 3 BaCl_{2} \rightarrow 6 NaCl + Ba_{3}(PO_{4})_{2}\] indicates that 2 moles of sodium phosphate react with 3 moles of barium chloride to produce 6 moles of sodium chloride and 1 mole of barium phosphate.

To achieve a successful reaction, one must ensure that the equation is balanced, which means the number of atoms for each element is the same on both sides of the equation. This reflects the Law of Conservation of Mass, where mass is neither created nor destroyed in a chemical reaction. A balanced equation provides a clear picture of the proportions needed for the reactants and the expected yield of each product, which is fundamental for the next steps in stoichiometry.

Limiting Reactant Determination

Limiting reactant determination is like finding out which ingredient will run out first in a recipe, limiting the amount of final product you can make. In chemical terms, the limiting reactant is the one that will be completely used up first in a reaction and thus determines the maximum amount of product that can be formed.

To identify the limiting reactant, we compare the number of moles of each reactant with the stoichiometry of the balanced equation. For our exercise, we calculated the moles of both sodium phosphate and barium chloride and then applied the stoichiometric ratios from the balanced equation. The comparison \[(0.100 \text{ moles Na}_3PO_4) / (2 \text{ moles Na}_3PO_4) = (0.1156 \text{ moles BaCl}_2) / (3 \text{ moles BaCl}_2)\] revealed that barium chloride is the limiting reactant.

This is critical information because it informs us how much product can be formed. In the context of our exercise, knowing which reactant is limiting allowed us to accurately calculate the moles of barium phosphate produced, leading us to find the mass of the precipitate. Without this step, you could either waste materials or end up with less product than expected.

Mole-to-Mass Conversion

Mole-to-mass conversion is the process of converting between the amount of substance (moles) and its corresponding mass. This is necessary because chemical reactions are often discussed in terms of moles, but practical applications require knowing the actual mass. The key to this conversion is the molar mass, which is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

In our exercise, the last step was to calculate the mass of barium phosphate formed. We did this by multiplying the moles of barium phosphate (0.03853) by its molar mass (601.98 g/mol), resulting in 23.2 g of barium phosphate. The molar mass is calculated by adding together the masses of all the atoms in the compound, considering their respective quantities.

The molar mass serves as a conversion factor between moles and grams, allowing chemists and students alike to bridge the gap between the microscopic (individual molecules and atoms) and the macroscopic (measurable amounts of chemicals). This conversion is vital in laboratory settings for measuring, purchasing, and applying substances with the precision that chemistry demands.