Problem 77 - Chemistry and Chemical Reactivity

Question

These questions are not designated as to type or location in the chapter. They may combine several concepts. Suppose \(16.04 \mathrm{g}\) of benzene, \(\mathbf{C}_{6} \mathbf{H}_{6},\) is burned in oxygen. (a) What are the products of the reaction? (b) Write a balanced equation for the reaction. (c) What mass of \(\mathbf{O}_{2},\) in grams, is required for complete combustion of benzene? (d) What is the total mass of products expected from the combustion of \(16.04 \mathrm{g}\) of benzene?

Step-by-Step Solution

Answer

The products are CO2 and H2O. 98.64 g of O2 is needed, and 114.68 g of products will form.

Step 1: Identify the Products

When benzene (\( \mathbf{C}_6\mathbf{H}_6 \)) undergoes complete combustion in oxygen, the primary products formed are carbon dioxide (\( \mathbf{CO}_2 \)) and water (\( \mathbf{H}_2\mathbf{O} \)).

Step 2: Write the Unbalanced Equation

The general equation for the combustion of benzene is:\[ \mathbf{C}_6\mathbf{H}_6 + \mathbf{O}_2 \rightarrow \mathbf{CO}_2 + \mathbf{H}_2\mathbf{O} \]

Step 3: Balance the Chemical Equation

To balance the equation, ensure the same number of each type of atom on both sides:\[ \mathbf{C}_6\mathbf{H}_6 + 15\mathbf{O}_2 \rightarrow 12\mathbf{CO}_2 + 6\mathbf{H}_2\mathbf{O} \]This balances carbon, hydrogen, and oxygen atoms.

Step 4: Calculate Molar Masses

Calculate the molar mass of benzene: \( \mathbf{C}_6\mathbf{H}_6 = 6(12) + 6(1) = 78 \text{ g/mol} \).Calculate the molar mass of oxygen: \( \mathbf{O}_2 = 2(16) = 32 \text{ g/mol} \).

Step 5: Moles of Benzene

Calculate the moles of benzene (\( \mathbf{C}_6\mathbf{H}_6 \)) using its mass and molar mass: \[ \text{Moles of } \mathbf{C}_6\mathbf{H}_6 = \frac{16.04 \text{ g}}{78 \text{ g/mol}} \approx 0.2055 \text{ mol} \]

Step 6: Calculate Moles of Oxygen Required

According to the balanced equation, each mole of benzene requires 15 moles of oxygen. So the moles of oxygen needed:\[ 0.2055 \text{ mol } \mathbf{C}_6\mathbf{H}_6 \times 15 \approx 3.0825 \text{ mol } \mathbf{O}_2 \]

Step 7: Calculate Mass of Oxygen Required

Convert moles of oxygen to grams using its molar mass:\[ \text{Mass of } \mathbf{O}_2 = 3.0825 \text{ mol} \times 32 \text{ g/mol} \approx 98.64 \text{ g} \]

Step 8: Total Mass of Products

The principle of conservation of mass requires that the mass of reactants equals the mass of products. Total mass of reactants = Total mass of products = (mass of benzene + mass of oxygen). Thus,\[ 16.04 \text{ g (benzene)} + 98.64 \text{ g (oxygen)} = 114.68 \text{ g (products)} \]

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is an essential step in solving reactions because it ensures that the Law of Conservation of Mass is respected. In a combustion reaction like the one between benzene ( C₆H₆ ) and oxygen ( O₂ ), we strive to have the same number of each atom type on both sides of the equation. This helps portray the reality that atoms are neither created nor destroyed in a reaction but merely rearranged.

Steps to Balance a Chemical Equation:

• Identify the reactants and products. This involves understanding what substances are reacting and what they form. In this case, benzene and oxygen are the reactants, while carbon dioxide and water are the products.
• Write the unbalanced chemical equation. For benzene's combustion, it begins as: C₆H₆ + O₂ → CO₂ + H₂O.
• Count the number of each type of atom on both sides of the equation.
• Adjust coefficients to ensure an equal number of each atom type on both sides. Here, the balanced equation becomes: C₆H₆ + 15O₂ → 12CO₂ + 6H₂O.
• Re-check to make sure all atoms are balanced, making any minor adjustments if necessary.

Balancing equations allows chemists to perform accurate stoichiometric calculations, informing how much of each chemical is needed or produced.

Stoichiometry Calculations

Stoichiometry involves calculating the proportions of reactants and products in chemical reactions. In the combustion of benzene, it's crucial to determine how much oxygen is required for the reaction to complete fully. This is achieved through stoichiometry calculations which utilize the balanced chemical equation.

Let's break down the steps:

• Identify the molar relationships from the balanced equation. For benzene's combustion, the balanced equation shows that 1 mole of C₆H₆ reacts with 15 moles of O₂.
• Use the known mass of benzene to find the moles of benzene. Using the formula: \[\frac{\text{mass of substance}}{\text{molar mass of substance}} = \text{moles of substance},\]we find the moles of benzene based on its mass and molar mass (78 g/mol).
• Using the mole ratio from the balanced equation, calculate the moles of oxygen required. Here, it is calculated as: \[0.2055 \text{ moles of } C₆H₆ \times \frac{15 \text{ moles of } O₂}{1 \text{ mole of } C₆H₆} \approx 3.0825 \text{ moles of } O₂.\]
• Convert moles of oxygen to grams using the molar mass of O₂ (32 g/mol) to find the necessary grams to complete the reaction.

Stoichiometry bridges the gap between the macroscopic world of grams and kilograms and the microscopic world of molecules and moles, offering precise calculations of materials used in reactions.

Molar Mass Determination

The molar mass of a compound is key to converting between mass and moles, a fundamental step in stoichiometry. Molar mass provides the bridge to calculate how much of a substance is present, whether in terms of grams, moles, or even molecules. Understanding the molar mass concept is crucial in accurately performing chemical calculations.

Determining Molar Mass:

• Identify the chemical formula of the compound. For benzene, this is C₆H₆.
• Locate the atomic masses of each element from the periodic table. Carbon (C) has an atomic mass of approximately 12 g/mol, and Hydrogen (H) about 1 g/mol.
• Multiply the atomic mass of each element by the number of atoms of that element in the formula. For benzene: \[6 \times 12 + 6 \times 1 = 78 \text{ g/mol}.\]
• Sum these values to obtain the molar mass of the compound.

This fundamental skill is not only critical when beginning with stoichiometry but is also applied across many aspects of chemical science, allowing for the precise calculation of reactants and products in chemical reactions.