Problem 76 - Chemistry: The Central Science

Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If \(7.50 \mathrm{~g}\) of sulfuric acid and \(7.50 \mathrm{~g}\) of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete.

Step-by-Step Solution

Answer

In the final mixture after the reaction is complete, there are \(5.23 \mathrm{~g}\) of sulfuric acid, no lead(II) acetate left (since it's the limiting reactant and has been completely consumed), \(6.99 \mathrm{~g}\) of lead(II) sulfate, and \(2.77 \mathrm{~g}\) of acetic acid.

Step 1: Write the balanced chemical equation

Write the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and lead(II) acetate (Pb(C2H3O2)2) forming lead(II) sulfate (PbSO4) and acetic acid (C2H4O2): \[H_{2}SO_{4} + Pb(C_{2}H_{3}O_{2})_{2} \rightarrow PbSO_{4} + 2C_{2}H_{4}O_{2}\]

Step 2: Calculate the moles of the reactants

Calculate the moles of sulfuric acid and lead(II) acetate using their molar masses: - Moles of sulfuric acid: \(7.50 \mathrm{g} \times \dfrac{1 \mathrm{~mol}}{98.08 \mathrm{~g/mol}} = 0.0764 \mathrm{~mol}\) - Moles of lead(II) acetate: \(7.50 \mathrm{g} \times \dfrac{1 \mathrm{~mol}}{325.3 \mathrm{~g/mol}} = 0.02305 \mathrm{~mol}\)

Step 3: Determine the limiting reactant

Compare the mole ratio of the reactants to the coefficients in the balanced equation to find the limiting reactant: - Sulfuric acid: \(0.0764 \mathrm{~mol} \times \dfrac{1}{1} = 0.0764\) - Lead(II) acetate: \(0.02305 \mathrm{~mol} \times \dfrac{1}{1} = 0.02305\) Since 0.02305 is smaller than 0.0764, lead(II) acetate is the limiting reactant.

Step 4: Calculate the moles of products formed

Use the stoichiometry of the balanced equation to calculate the moles of lead(II) sulfate and acetic acid formed: - Moles of lead(II) sulfate: \(0.02305 \mathrm{~mol} \times \dfrac{1}{1} = 0.02305 \mathrm{~mol}\) - Moles of acetic acid: \(0.02305 \mathrm{~mol} \times \dfrac{2}{1} = 0.0461 \mathrm{~mol}\)

Step 5: Calculate the moles of unreacted sulfuric acid

Calculate the moles of unreacted sulfuric acid, considering that 0.02305 mol of it has reacted: Unreacted moles of sulfuric acid: \(0.0764 \mathrm{~mol} - 0.02305 \mathrm{~mol} = 0.05335 \mathrm{~mol}\)

Step 6: Calculate the mass of products and unreacted sulfuric acid

Convert the moles of each substance to grams using their molar masses: - Mass of unreacted sulfuric acid: \(0.05335 \mathrm{~mol} \times 98.08 \mathrm{~g/mol} = 5.23 \mathrm{~g}\) - Mass of lead(II) sulfate: \(0.02305 \mathrm{~mol} \times 303.3 \mathrm{~g/mol} = 6.99 \mathrm{~g}\) - Mass of acetic acid: \(0.0461 \mathrm{~mol} \times 60.05 \mathrm{~g/mol} = 2.77 \mathrm{~g}\) In the final mixture after the reaction is complete, there are: - \(5.23 \mathrm{~g}\) of sulfuric acid - No lead(II) acetate left (since it's the limiting reactant and has been completely consumed) - \(6.99 \mathrm{~g}\) of lead(II) sulfate - \(2.77 \mathrm{~g}\) of acetic acid

Key Concepts

Chemical Reaction

A chemical reaction is a process where substances, called reactants, transform into entirely new substances called products. During a chemical reaction, the atoms of the reactants rearrange themselves to create different molecule combinations.

For example, when sulfuric acid reacts with lead(II) acetate, as described in our exercise, the acid donates a proton to the acetate, which triggers a series of bond rearrangements. This ultimately results in the formation of lead(II) sulfate and acetic acid. It's like a dance of particles where some decide to change partners, forming new relationships that are chemically stable as products.

Limiting Reactant

In stoichiometry, the limiting reactant is the substance that's completely used up first, restricting the amount of product formed. Imagine you're making sandwiches and you have plenty of bread but only a few slices of cheese. The cheese is your limiting reactant because it runs out first, and without it, no more sandwiches can be made, regardless of how much bread you have left.

Identifying the limiting reactant involves comparing the amounts of reactants used in a chemical reaction with the reactants' stoichiometric coefficients from the balanced chemical equation. In the sulfuric acid and lead(II) acetate reaction, the limiting reactant is determined based on which compound produces the least amount of product—it dictates the extent of the reaction.

Balanced Chemical Equation

A balanced chemical equation is a representation of a chemical reaction that includes the number of atoms for each element in the reactants and the products, and conserves the same amount of those atoms according to the Law of Conservation of Mass. Think of it as a recipe that tells you exactly how much of each ingredient you need to make a perfect dish.

For the reaction between sulfuric acid and lead(II) acetate, the balanced equation is necessary because it tells us the mole ratio in which reactants combine to form products. This ratio is used to deduce the number of moles and ultimately the mass of each substance involved in the reaction after completion.

Molar Mass

Molar mass is the weight of one mole (Avogadro's number of atoms or molecules) of a substance and is expressed in grams per mole (g/mol). In simpler terms, it's like knowing the weight of a dozen of different fruits. A dozen bananas weigh more than a dozen cherries. This concept helps chemists relate the mass of a substance to the number of particles it contains.

Using the molar mass, you can convert the grams of a substance to moles, which is a key step in stoichiometry. In our example, we use the molar masses of sulfuric acid and lead(II) acetate to find out how many moles of each are present initially, and then use the molar mass again to convert the moles of the products and remaining reactants back into grams in the final calculation.