Question

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

Short answer

The mass of the solid barium phosphate produced when a 500.0 mL sample of 0.200 M sodium phosphate is mixed with 400.0 mL of 0.289 M barium chloride is approximately 23.17 grams.

Step-by-step solution

Write the balanced chemical equation for the reaction:

Combining sodium phosphate (Na3PO4) with barium chloride (BaCl2) will produce barium phosphate (Ba3(PO4)2) and sodium chloride (NaCl). The balanced chemical equation for the reaction is: \(3BaCl_{2(aq)} + 2Na_{3}PO_{4(aq)} \rightarrow Ba_{3}(PO_{4})_{2(s)} + 6NaCl_{(aq)}\) The solid produced in this reaction is barium phosphate, Ba3(PO4)2.

Determine the limiting reactant:

We have to determine which reactant is consumed completely during the reaction, known as the limiting reactant. We know the initial volumes and concentrations of both reactants: Sodium phosphate: Volume = 500 mL, Concentration = 0.200 M Barium chloride: Volume = 400 mL, Concentration = 0.289 M Next, let's calculate the initial moles of each reactant: Moles of sodium phosphate = Volume × Concentration = 500 mL × 0.200 M = 100 mmol Moles of barium chloride = Volume × Concentration = 400 mL × 0.289 M = 115.6 mmol Now, we will compare the mole ratio of the reactants to the stoichiometric ratio in the balanced equation: Mole ratio = (Moles Na3PO4) / (Moles BaCl2) = 100 mmol / 115.6 mmol ≈ 0.865 Stoichiometric ratio = 2 Na3PO4 / 3 BaCl2 = 2/3 ≈ 0.667 Since the mole ratio (0.865) is greater than the stoichiometric ratio (0.667), sodium phosphate is in excess and barium chloride is the limiting reactant.

Calculate the moles of barium phosphate produced:

According to the balanced chemical equation, 3 moles of barium chloride react with 2 moles of sodium phosphate to form 1 mole of barium phosphate. Since we know that barium chloride is the limiting reactant, we can calculate how much barium phosphate is produced: Moles of barium phosphate = (Moles of BaCl2) × (1 mole Ba3(PO4)2 / 3 moles BaCl2) Moles of barium phosphate = 115.6 mmol × (1/3) ≈ 38.53 mmol

Calculate the mass of barium phosphate produced:

To find the mass of the solid produced, we need the molar mass of barium phosphate: Molar Mass of Ba3(PO4)2 = (3 × Ba) + (2 × 3 × P) + (2 × 4 × O) Molar Mass of Ba3(PO4)2 = (3 × 137.33 g/mol) + (2 × 3 × 30.97 g/mol) + (2 × 4 × 16.00 g/mol) Molar Mass of Ba3(PO4)2 ≈ 601.93 g/mol Now, we can find the mass of barium phosphate produced: Mass of Ba3(PO4)2 = Moles of Ba3(PO4)2 × Molar Mass of Ba3(PO4)2 Mass of Ba3(PO4)2 ≈ 38.53 mmol × 601.93 g/mol ≈ 23.17 g Thus, the mass of the solid produced (barium phosphate) is approximately 23.17 grams.

Key concepts

Limiting Reactant

In the world of chemistry, when a reaction occurs, one of the reactants is usually used up before the others. This reactant is known as the "limiting reactant." It dictates the maximum amount of product that the reaction can produce because once it is completely consumed, the reaction cannot proceed further.
To determine which reactant is the limiting one, chemists compare the initial moles of each reactant to the stoichiometric ratios derived from the balanced chemical equation.

• Calculate the moles of each reactant available initially.
• Use the coefficients from the balanced equation to determine the ratio needed for the reaction to occur completely.
• Compare these ratios to find which reactant would be consumed fastest. This reactant is your limiting reactant.

In our example, after determining the moles of sodium phosphate and barium chloride available, we found barium chloride to be the limiting reactant because it would run out first based on the stoichiometric ratios.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants (starting materials) on the left and the products (resulting substances) on the right. The balanced equation ensures that the number of atoms for each element is the same on both sides, adhering to the law of conservation of mass.
For the reaction between sodium phosphate and barium chloride, we write:

• Reactants: Sodium phosphate (\( Na_3PO_4 \)) and Barium chloride (\( BaCl_2 \)).
• Products: Barium phosphate (\( Ba_3(PO_4)_2 \), a solid) and Sodium chloride (\( NaCl \)).

The equation becomes:\[ 3BaCl_{2(aq)} + 2Na_{3}PO_{4(aq)} \rightarrow Ba_{3}(PO_{4})_{2(s)} + 6NaCl_{(aq)} \]This notation provides a clear guide on the ratios in which reactants react and products form, which is crucial for understanding how much of each substance is involved in the reaction.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It allows us to convert between the number of moles of a substance and its mass, essential for calculating the quantities of reactants and products involved in a chemical reaction.
To determine the molar mass, you sum the atomic masses of all atoms in a molecule. The atomic masses are typically found on the periodic table.
For barium phosphate (\( Ba_3(PO_4)_2 \)), the molar mass is calculated as follows:

• Barium (\( Ba \)): 137.33 g/mol
• Phosphorus (\( P \)): 30.97 g/mol
• Oxygen (\( O \)): 16.00 g/mol

Calculate the molar mass:\[ (3 \, \times \, 137.33) + (2 \, \times \, 3 \, \times \, 30.97) + (2 \, \times \, 4 \, \times \, 16.00) = 601.93 \, g/mol \]Knowing this molar mass allows us to convert the moles of barium phosphate obtained from the reaction into a mass, which gives us a tangible measure of the product formed.