Problem 79 - Chemistry and Chemical Reactivity

Question

Suppose \(16.04 \mathrm{g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) is burned in oxygen. (a) What are the products of the reaction? (b) Write a balanced equation for the reaction. (c) What mass of \(\mathrm{O}_{2}\), in grams, is required for complete combustion of benzene? (d) What is the total mass of products expected from the combustion of 16.04 g of benzene?

Step-by-Step Solution

Answer

The products are CO₂ and H₂O. 49.27 g of O₂ is needed, producing 65.31 g of total products.

Step 1: Identify the Reaction Products

When benzene (\(\mathrm{C}_6\mathrm{H}_6\)) is burned in oxygen (\(\mathrm{O}_2\)), the main products are carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). This is a standard combustion reaction for hydrocarbons.

Step 2: Write the Balanced Combustion Equation

Write and balance the chemical equation: \(\mathrm{C}_6\mathrm{H}_6 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\). Balancing this gives you: \(2\mathrm{C}_6\mathrm{H}_6 + 15\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}\).

Step 3: Calculate Molar Mass of Benzene

The molar mass of benzene (\(\mathrm{C}_6\mathrm{H}_6\)) is calculated as: \(6 \times 12.01 \text{ (C)} + 6 \times 1.01 \text{ (H)} = 78.12\, \mathrm{g/mol}\).

Step 4: Determine Moles of Benzene

Using the given mass of benzene, find moles: \[\text{moles of } \mathrm{C}_6\mathrm{H}_6 = \frac{16.04 \text{ g}}{78.12\, \mathrm{g/mol}} \approx 0.2053\, \mathrm{mol}\].

Step 5: Calculate Required Oxygen Moles

From the balanced equation, \(2\mathrm{C}_6\mathrm{H}_6\) requires \(15\mathrm{O}_2\), so 1 mole of \(\mathrm{C}_6\mathrm{H}_6\) requires \(7.5\mathrm{O}_2\). Therefore, \(0.2053\, \, \mathrm{mol}\) of benzene needs \((0.2053) \times 7.5 = 1.5397 \, \mathrm{mol} \, \mathrm{O}_2\).

Step 6: Convert Moles of Oxygen to Mass

Molar mass of \(\mathrm{O}_2\) is \(32.00\, \mathrm{g/mol}\). Thus, the mass of \(\mathrm{O}_2\) is: \[\text{mass of } \mathrm{O}_2 = 1.5397 \, \mathrm{mol} \times 32.00\, \mathrm{g/mol} \approx 49.27 \text{ g}\].

Step 7: Calculate Total Mass of Products

The total mass of products equals the sum of the masses of reactants (law of conservation of mass). Combine the mass of benzene (16.04 g) with the mass of oxygen (49.27 g) to calculate: \(16.04 \, \mathrm{g} + 49.27 \, \mathrm{g} = 65.31 \, \mathrm{g}\).

Key Concepts

Chemical Equation Balancing

Balancing a chemical equation is like solving a puzzle to ensure that the same number of each type of atom appears on both sides of the equation. This means you must adjust the coefficients, the numbers in front of the molecules, without altering the chemical formulas themselves.
In the combustion reaction of benzene (C₆H₆), we aim to produce carbon dioxide (CO₂) and water (H₂O).

• Start by balancing the carbon atoms: C₆H₆ has six carbon atoms, so you need six CO₂ molecules on the product side, resulting in 12 carbon atoms in total.
• Next, balance the hydrogen atoms by ensuring that for six hydrogen atoms in C₆H₆, we use three H₂O molecules, providing a total of six hydrogen atoms.
• Finally, balance the oxygen atoms, which requires 15 O₂ molecules on the reactant side to meet the needs of forming 12 CO₂ and 6 H₂O molecules.

The balanced equation is:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O.

Mass Calculation

Mass calculation ensures that the total mass of reactants equals the total mass of products according to the law of conservation of mass. When we calculate the mass of oxygen required for combustion, we use the molar mass of the substances involved.
The calculated mass of O₂ comes from multiplying its molar quantity by its molar mass:

• Mass of benzene = 16.04 g
• Mass of oxygen calculated in moles = 1.5397 mol
• Moles to mass conversion: 1.5397 mol × 32.00 g/mol = 49.27 g

Knowing these values helps ensure the reaction observes the conservation of mass, turning reactants into equivalent product mass.

Molar Mass

Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It's essential for converting between mass and moles.
To find the molar mass of benzene (C₆H₆), we calculate the sum of the atomic masses of its constituent atoms:

• Carbon: 6 atoms × 12.01 g/mol = 72.06 g/mol
• Hydrogen: 6 atoms × 1.01 g/mol = 6.06 g/mol

Therefore, the molar mass of benzene is 78.12 g/mol.
This allows us to convert the given mass of benzene into moles, providing a bridge to stoichiometric calculations.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It's crucial for determining how much of each chemical is needed or produced.
In the benzene combustion reaction, the stoichiometry is derived from the balanced equation. We see that:

• 2 moles of benzene need 15 moles of oxygen to react completely.
• For every 1 mole of benzene, 7.5 moles of oxygen are required.

Using the known moles of benzene (0.2053 mol), we can find the required moles of oxygen by multiplying it by the stoichiometric ratio (7.5), resulting in approximately 1.5397 mol of O₂.
This stoichiometry calculation helps predict the mass and volume of reactants or products in any given reaction.