Problem 82 - 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 5.00 g of sulfuric acid and 5.00 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

After the reaction is complete, there are 3.49 g of sulfuric acid, 0 g of lead(II) acetate, 4.67 g of lead(II) sulfate, and 1.85 g of acetic acid present in the mixture.

Step 1: Write a balanced chemical equation

The balanced chemical equation for the reaction between sulfuric acid and lead(II) acetate is: \[H_2SO_4 + Pb(C_2H_3O_2)_2 \rightarrow PbSO_4\downarrow + 2C_2H_4O_2\]

Step 2: Determine the limiting reagent

We have 5.00 g of sulfuric acid and 5.00 g of lead(II) acetate. We determine the number of moles of each of them using their molar masses (98.08 g/mol for H2SO4 and 325.29 g/mol for Pb(C2H3O2)2): Moles of H2SO4 = (5.00 g) / (98.08 g/mol) = 0.0510 moles Moles of Pb(C2H3O2)2 = (5.00 g) / (325.29 g/mol) = 0.0154 moles Now we will determine which compound is the limiting reagent by comparing their mole ratios, according to the balanced chemical equation: 0.0510 moles of H2SO4 / 1 = 0.0510 0.0154 moles of Pb(C2H3O2)2 / 1 = 0.0154 Since 0.0154 is lower than 0.0510, the limiting reagent is lead(II) acetate.

Step 3: Calculate the number of moles reacted and produced

From the balanced chemical equation, we can see that for each mole of Pb(C2H3O2)2 that reacts, one mole of PbSO4 and two moles of C2H4O2 are produced: Moles of PbSO4 produced = 0.0154 moles Moles of C2H4O2 produced = 0.0154 moles × 2 = 0.0308 moles Now we can determine the number of moles of H2SO4 remaining: Moles of H2SO4 reacted = 0.0154 moles Moles of H2SO4 remaining = 0.0510 moles - 0.0154 moles = 0.0356 moles

Step 4: Calculate the mass of the products and reactants after the reaction

Using the molar masses of each compound (98.08 g/mol for H2SO4, 325.29 g/mol for Pb(C2H3O2)2, 303.27 g/mol for PbSO4, and 60.05 g/mol for C2H4O2), we can calculate the masses after the reaction: Mass of H2SO4 = 0.0356 moles × 98.08 g/mol = 3.49 g Mass of Pb(C2H3O2)2 (completely reacted) = 0 g Mass of PbSO4 = 0.0154 moles × 303.27 g/mol = 4.67 g Mass of C2H4O2 = 0.0308 moles × 60.05 g/mol = 1.85 g So after the reaction is complete, there are 3.49 g of sulfuric acid, 0 g of lead(II) acetate, 4.67 g of lead(II) sulfate, and 1.85 g of acetic acid present in the mixture.

Key Concepts

Limiting Reagent

Imagine you're baking cookies, and you have a limited amount of chocolate chips. No matter how much of the other ingredients you have, you can only make as many cookies as your chocolate chips allow. That's how a limiting reagent works in chemistry. It's the ingredient (or reagent) in a chemical reaction that runs out first, limiting the amount of product that can be formed.

In our exercise, we identified lead(II) acetate as the limiting reagent by comparing its mole quantity to that of sulfuric acid. By finding out which reagent has the smallest mole-to-coefficient ratio based on the balanced chemical equation, we can infer which substance will entirely be consumed first, dictating the extent of the reaction. This step is vital because it not only tells us how much product will form but also helps in calculating the amount of excess reagent left over after the reaction.

Balanced Chemical Equation

A balanced chemical equation is like a recipe for a chemical reaction; just as recipes list exact amounts of each ingredient, a balanced chemical equation lists the exact amount of reactants needed to create certain products without any leftovers. It adheres to the conservation of mass: the idea that matter cannot be created or destroyed, only transformed.

In our example, the equation \(H_2SO_4 + Pb(C_2H_3O_2)_2 \rightarrow PbSO_4\downarrow + 2C_2H_4O_2\) reveals exactly how many molecules of each substance interact to form the products. This equation is crucial because it allows us to perform calculations that predict the amounts of substances consumed and created during the reaction, providing critical information for both scientific understanding and practical applications, such as manufacturing and laboratory work.

Mole-to-Mass Conversion

Chemists often deal with the mole, which is a unit that speaks to the quantity of a substance based on the number of particles it contains. But in the real world, we measure chemicals by weight, so converting moles to grams is a common task in chemistry known as mole-to-mass conversion.

To perform this conversion, you need to know the molar mass of the substance, which tells you how much one mole of that substance weighs. In our exercise, we used the molar masses of sulfuric acid and lead(II) acetate to convert grams to moles. This step was crucial for finding the limiting reagent. Additionally, after knowing which products form and how many moles of them we expect (from the balanced chemical equation and the limiting reagent), we converted moles back to grams to determine the mass of each product. This conversion is essential because it provides a bridge between the microscopic scale of atoms and molecules and the macroscopic world we experience.