Question

How many grams of barium sulfate form when \(35.0 \mathrm{~mL}\) of \(0.160 \mathrm{M}\) barium chloride reacts with \(58.0 \mathrm{~mL}\) of \(0.065 \mathrm{M}\) sodium sulfate?

Short answer

0.880 grams of barium sulfate are formed.

Step-by-step solution

Write the balanced chemical equation

The reaction between barium chloride (BaCl2) and sodium sulfate (Na2SO4) forms barium sulfate (BaSO4) and sodium chloride (NaCl). The balanced chemical equation is: \[ \text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow \text{BaSO}_4 (s) + 2 \text{NaCl} (aq) \]

Calculate the moles of barium chloride

The number of moles of barium chloride (BaCl2) is calculated using its volume and molarity:\[ \text{Moles of BaCl}_2 = 0.160 \text{ M} \times 0.0350 \text{ L} = 0.00560 \text{ moles} \]

Calculate the moles of sodium sulfate

The number of moles of sodium sulfate (Na2SO4) is calculated using its volume and molarity: \[ \text{Moles of Na}_2\text{SO}_4 = 0.065 \text{ M} \times 0.058 \text{ L} = 0.00377 \text{ moles} \]

Determine the limiting reagent

Compare the stoichiometric ratio from the balanced equation to determine the limiting reagent. The balanced equation shows a 1:1 molar ratio between BaCl2 and Na2SO4. The reagent with fewer moles is the limiting reagent. In this case, sodium sulfate (0.00377 moles) is the limiting reagent.

Calculate the moles of barium sulfate formed

Using the moles of the limiting reagent (sodium sulfate), determine the moles of barium sulfate formed. According to the stoichiometric relationship, 0.00377 moles of Na2SO4 produces 0.00377 moles of BaSO4.

Calculate the mass of barium sulfate formed

The molar mass of barium sulfate (BaSO4) is calculated from its atomic masses:\[ \text{Molar mass of BaSO}_4 = \text{Ba} (137.33) + \text{S} (32.07) + 4 \times \text{O} (16.00) = 233.40 \text{ g/mol} \] The mass of barium sulfate is then:\[ \text{Mass of BaSO}_4 = 0.00377 \text{ moles} \times 233.40 \text{ g/mol} = 0.880 \text{ grams} \]

Key concepts

Molarity Calculation

Molarity is a measure of the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity (M) is given by \ \( M = \frac{n}{V} \ \) where n is the number of moles of solute and V is the volume of the solution in liters.
For example, in this exercise, to find the moles of barium chloride (BaCl2), we used the molarity (0.160 M) and the volume (35.0 mL or 0.035 L):

\( \text{Moles of BaCl}_2 = 0.160 \text{ M} \times 0.0350 \text{ L} = 0.00560 \text{ moles} \ \)

This approach helps us determine how much of a substance is present in a given volume of solution.

Limiting Reagent

The limiting reagent in a chemical reaction is the substance that gets used up first, thus determining the amount of product formed. To identify the limiting reagent, compare the moles of each reactant based on the stoichiometric coefficients from the balanced chemical equation.
In the given exercise, the balanced equation is:

\( \text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow \text{BaSO}_4 (s) + 2 \text{NaCl} (aq) \ \)

The stoichiometric ratio between BaCl2 and Na2SO4 is 1:1. We calculated the moles of each reactant and found that we had 0.00560 moles of BaCl2 and 0.00377 moles of Na2SO4. Since Na2SO4 has fewer moles, it is the limiting reagent.

Chemical Reaction Equations

Chemical reaction equations represent the substances involved in a reaction and their proportions. A balanced chemical equation ensures that the number of atoms of each element is the same on both sides of the equation, which corresponds to the law of conservation of mass.
In our example, the balanced equation for the reaction between barium chloride and sodium sulfate is:

\( \text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow \text{BaSO}_4 (s) + 2 \text{NaCl} (aq) \ \)

This equation shows that one mole of BaCl2 reacts with one mole of Na2SO4 to produce one mole of BaSO4 and two moles of NaCl. Balancing chemical equations is crucial for correctly calculating reactants and products in stoichiometry problems.

Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. One mole contains exactly \( 6.022 \times 10^{23} \) particles (Avogadro's number). The mole concept helps in converting between the mass of a substance and the number of particles or volume of a gas.
In the exercise, we calculated moles of reactants using their molarity and volume. For instance:

\( \text{Moles of BaCl}_2 = 0.160 \text{ M} \times 0.0350 \text{ L} = 0.00560 \text{ moles} \ \)

And similarly for Na2SO4:

\( \text{Moles of Na}_2\text{SO}_4 = 0.065 \text{ M} \times 0.058 \text{ L} = 0.00377 \text{ moles} \ \)

Understanding the mole concept is essential for stoichiometry calculations and determining reactants and products in chemical reactions.