Problem 81 - Chemistry Matter and Change

Question

Alkaline Battery An alkaline battery produces electrical energy according to this equation. \begin{equation} \mathrm{Zn}(\mathrm{s})+2 \mathrm{MnO}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Mn}_{2} \mathrm{O}_{3}(\mathrm{s}) \end{equation} \begin{equation} \begin{array}{l}{\text { a. Determine the limiting reactant if } 25.0 \mathrm{g} \text { of } \mathrm{Zn} \text { and }} \\ {30.0 \mathrm{g} \text { of } \mathrm{MnO}_{2} \text { are used. }} \\ {\text { b. Determine the mass of } \mathrm{Zn}(\mathrm{OH})_{2} \text { produced. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

MnO₂ is the limiting reactant, and 17.15 g of Zn(OH)₂ is produced.

Step 1: Calculate Moles of Zn

First, determine the number of moles of Zn used. The molar mass of Zn is approximately 65.38 g/mol. Use the formula: \[\text{moles of Zn} = \frac{25.0 \text{ g}}{65.38 \text{ g/mol}} \approx 0.382 \text{ mol} \]

Step 2: Calculate Moles of MnO2

Next, determine the number of moles of \( \text{MnO}_2 \). The molar mass of \( \text{MnO}_2 \) is approximately 86.94 g/mol. Calculate using:\[\text{moles of } \text{MnO}_2 = \frac{30.0 \text{ g}}{86.94 \text{ g/mol}} \approx 0.345 \text{ mol} \]

Step 3: Determine Limiting Reactant

The balanced chemical equation indicates a 1:2 molar ratio between Zn and MnO₂. For 0.382 mol Zn, we need 0.764 mol of MnO₂. We only have 0.345 mol of MnO₂, making MnO₂ the limiting reactant.

Step 4: Calculate Moles of Zn(OH)2 Produced

According to the equation, 2 moles of MnO₂ produce 1 mole of Zn(OH)₂. Therefore, 0.345 mol of MnO₂ will produce:\[\text{moles of Zn(OH)}_2 = \frac{0.345 \text{ mol MnO}_2}{2} = 0.1725 \text{ mol Zn(OH)}_2 \]

Step 5: Calculate Mass of Zn(OH)2 Produced

Finally, convert moles of Zn(OH)₂ to mass. Zn(OH)₂ has a molar mass of approximately 99.42 g/mol:\[\text{mass of Zn(OH)}_2 = 0.1725 \text{ mol} \times 99.42 \text{ g/mol} \approx 17.15 \text{ g} \]

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is entirely consumed first, limiting the amount of products formed. To determine the limiting reactant, compare the mole ratio between the reactants used with the ratio presented in the balanced equation. For example, in the alkaline battery reaction involving Zn and MnO\(_2\), we examine their amounts in moles relative to the stoichiometric ratios.

The equation shows a 1:2 relationship between Zn and MnO\(_2\), meaning each mole of Zn requires two moles of MnO\(_2\). Since there are 0.382 moles of Zn and only 0.345 moles of MnO\(_2\) available, we quickly observe that MnO\(_2\) is the limiting reactant, as it runs out before there is enough to react with all of the Zn. This concept is crucial as it dictates the maximum amount of product that can be formed.

Chemical Equations

A chemical equation represents what happens during a chemical reaction. It displays the reactants and products along with their respective coefficients, which indicate the relative amounts in moles. This balanced representation is necessary to understand how molecules interact with one another in a chemical process.

In our example, the alkaline battery’s chemical equation involving Zn and MnO\(_2\) includes water and forms Zn(OH)\(_2\) and Mn\(_2\)O\(_3\). Balancing this equation is key, allowing us to derive the stoichiometric ratios, which we later utilize to determine the limiting reactant and calculate the products. A balanced equation ensures that the law of conservation of mass is maintained, meaning the number of atoms for each element is equal on both sides of the equation.

• Zn: 1
• MnO\(_2\): 2
• H\(_2\)O: 1
• Zn(OH)\(_2\): 1
• Mn\(_2\)O\(_3\): 1

Understanding chemical equations enables us to predict the quantities of products formed and manage resources efficiently.

Mole Calculations

Mole calculations provide a bridge between the microscopic world of atoms and the macroscopic world we can measure. They allow us to quantify substances involved in chemical reactions using their mass and molar mass. Here’s how to do mole calculations in practice:

- **Identify Molar Mass:** Each element or compound has a specific molar mass found on the periodic table. For Zn, the molar mass is approximately 65.38 g/mol, and for MnO\(_2\), it is roughly 86.94 g/mol.
- **Calculate Moles:** Use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \) to determine the number of moles. This step is essential for comparisons and limiting reactant identification.

For instance, with 25.0 grams of Zn, we calculate 0.382 moles, and with 30.0 grams of MnO\(_2\), we figure out 0.345 moles. From such calculations, we can then derive how much product will be produced and confirm which reactant limits the process. Mastering mole calculations ensures precision in evaluating and predicting outcomes in chemical reactions.