Question

Car Battery Car batteries use lead, lead(IV) oxide, and a sulfuric acid solution to produce an electric current. The products of the reaction are lead(II) sulfate in solution and water. a. Write the balanced equation for the reaction. b. Determine the mass of lead(II) sulfate produced when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid.

Short answer

a. The balanced chemical equation is: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O b. The mass of lead(II) sulfate produced when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid is 73.17 g.

Step-by-step solution

Identify the chemical formulas of the reactants and products.

For the reactants, we have: - Lead (Pb) - Lead (IV) oxide (PbO2) - Sulfuric acid (H2SO4) For the products, we have: - Lead(II) sulfate (PbSO4) - Water (H2O)

Write the unbalanced chemical equation.

The unbalanced chemical equation would be: Pb + PbO2 + H2SO4 → PbSO4 + H2O

Balance the chemical equation.

Balancing the chemical equation, we get: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O So, the balanced chemical equation is: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O b. Determine the mass of lead(II) sulfate produced:

Calculate the molar mass of lead(II) sulfate, lead, lead(IV) oxide and sulfuric acid.

The molar mass of the compounds are as follows: - Pb (lead): 207.2 g/mol - PbO2 (lead (IV) oxide): 239.2 g/mol - H2SO4 (sulfuric acid): 2(1.0) + 32.1 + 4(16.0) = 98.1 g/mol - PbSO4 (lead(II) sulfate): 207.2 + 32.1 + 4(16.0) = 303.3 g/mol

Calculate the moles of lead reacted.

We were given 25.0 g of lead. To calculate the moles of lead reacted, we can use the formula: moles = mass / molar mass moles of Pb = 25.0 g / 207.2 g/mol = 0.1207 mol

Determine the stoichiometric ratio

From the balanced chemical equation, Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O, the stoichiometric ratio between Pb and PbSO4 is 1:2.

Calculate the moles of lead(II) sulfate produced.

Since the stoichiometric ratio of lead to lead(II) sulfate is 1:2, for every mole of lead reacted, two moles of lead(II) sulfate are produced. Therefore, the moles of PbSO4 produced will be: moles of PbSO4 = 2 × moles of Pb = 2 × 0.1207 mol = 0.2414 mol

Calculate the mass of lead(II) sulfate produced.

Now we will use the molar mass of lead(II) sulfate to find out the mass of PbSO4 produced: mass of PbSO4 = moles of PbSO4 × molar mass of PbSO4 = 0.2414 mol × 303.3 g/mol = 73.17 g So, the mass of lead(II) sulfate produced when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid is 73.17 g.

Key concepts

Balanced Chemical Equations

When a chemical reaction occurs, the atoms involved are neither created nor destroyed; they simply rearrange to form new substances. The Law of Conservation of Mass tells us that we must have the same number of atoms of each element on both sides of the equation. This principle is vital when we write balanced chemical equations.

In the case of the reaction taking place in a car battery, where lead, lead(IV) oxide, and sulfuric acid react to produce lead(II) sulfate and water, the balanced chemical equation is essential to understanding the reaction. To achieve a balanced equation, we adjust the coefficients, which are the numbers in front of the chemical formulas. These coefficients indicate the ratios of the reactants and products in the reaction.

For instance, in the provided solution, the equation is balanced by placing the coefficient '2' before both H₂SO₄ and PbSO₄, as well as before H₂O, which ensures that there are equal numbers of each type of atom on both sides of the equation, conforming with the Law of Conservation of Mass.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves calculating the quantities of reactants and products involved in a chemical reaction. It is based on the balanced chemical equation and the concept of the mole, which is a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles.

In the battery reaction example, stoichiometry allows us to understand the relationship between reactants and products. From the balanced chemical equation, we could deduce that one mole of lead reacts with one mole of lead(IV) oxide and two moles of sulfuric acid to produce two moles of lead(II) sulfate and two moles of water. This stoichiometric ratio is crucial when predicting how much product is formed from a certain amount of reactant, or conversely, how much reactant is needed to produce a desired amount of product.

Molar Mass Calculations

The concept of molar mass is a pivotal element of stoichiometry, intersecting with both balanced chemical equations and the conversions of mass to moles and vice versa. Molar mass is defined as the mass of one mole of a substance, and it is usually expressed in grams per mole (g/mol).

To perform molar mass calculations, as seen in the battery reaction, each element's atomic mass must be taken from the periodic table, then multiplied by the number of atoms of that element in the compound and summed to find the molar mass of the entire compound. For example, in our reaction, we calculated the molar mass of lead(II) sulfate (PbSO₄) by adding the atomic masses of lead, sulfur, and four oxygen atoms. These calculations are necessary for translating the amount of a substance from grams, a unit of mass, to moles, a unit that relates to the number of particles in a chemical reaction.