Problem 47 - 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) What is the 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 \(16.04 \mathrm{g}\) of benzene?

Step-by-Step Solution

Answer

(a) CO₂ and H₂O; (b) 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O; (c) 49.22 g O₂; (d) 65.26 g products.

Step 1: Determine the Products

When benzene \((\text{C}_6\text{H}_6)\) is burned in oxygen \((\text{O}_2)\), it undergoes complete combustion, producing carbon dioxide \((\text{CO}_2)\) and water \((\text{H}_2\text{O})\) as products. Therefore, the products of the reaction are carbon dioxide and water.

Step 2: Write the Chemical Equation

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

Step 3: Balance the Equation

Balance the chemical equation for the combustion of benzene:1. Balance carbon: \( \text{C}_6\text{H}_6 \to 6 \text{CO}_2 \)2. Balance hydrogen: \( \text{H}_6 \to 3 \text{H}_2\text{O} \)3. Balance oxygen: \( 6 \text{CO}_2 + 3 \text{H}_2\text{O} \) needs 15 \( \text{O} \) atoms, so you need 7.5 \( \text{O}_2 \) molecules.Thus, the balanced equation is:\[ \text{2C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12 \text{CO}_2 + 6\text{H}_2\text{O} \]

Step 4: Calculate Moles of Benzene

Find the number of moles of benzene using its molar mass:- Molar mass of \( \text{C}_6\text{H}_6 \) is approximately 78.11 g/mol.- Moles of benzene = \( \frac{16.04}{78.11} \approx 0.205 \text{ mol} \).

Step 5: Calculate Required Moles of Oxygen

From the balanced equation, \( 2 \text{C}_6\text{H}_6 \) requires \( 15 \text{O}_2 \) molecules:- Therefore, \( 0.205 \text{ mol} \) of benzene requires \( 0.205 \times 7.5 \approx 1.538 \text{ mol} \) of \( \text{O}_2 \).

Step 6: Convert Oxygen Moles to Grams

Convert moles of \( \text{O}_2 \) to grams using its molar mass (32.00 g/mol):- Mass of \( \text{O}_2 = 1.538 \times 32.00 = 49.22 \text{ g} \).

Step 7: Calculate Total Mass of Products

From the law of conservation of mass, the mass of reactants equals the mass of products:- Total mass of products = mass of benzene + mass of \( \text{O}_2 \).- Total mass of products = \( 16.04 + 49.22 = 65.26 \text{ g} \).

Key Concepts

Chemical Equation

A chemical equation is a symbolic representation that portrays a chemical reaction in terms of symbols and formulas. It concisely communicates what reactants are consumed and what products are formed during the process. For example, when benzene
(\(\text{C}_6\text{H}_6\)) is burned in oxygen (\(\text{O}_2\)), the chemical equation reads as:\[\text{C}_6\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\]This equation shows benzene reacts with oxygen to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).

• The left side of the equation (reactants) represents benzene and oxygen.
• The right side (products) shows the resulting carbon dioxide and water.

Understanding chemical equations is crucial since they indicate how different substances interact during a reaction.

Balancing Equations

Balancing chemical equations is a necessary step to ensure the conservation of atoms from reactants to products. The law of conservation of mass requires that the mass and number of atoms in the reactants must equal those in the products. In the combustion of benzene, the unbalanced equation was:\[\text{C}_6\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\]To balance it, we need equal numbers of each type of atom on both sides.

• Balance carbon: 6 carbon atoms in benzene give 6 \( \text{CO}_2 \) molecules.
• Balance hydrogen: 6 hydrogen atoms require 3 water molecules ( \( 3\text{H}_2\text{O} \)).
• Balance oxygen: From 6 \( \text{CO}_2 \) and 3 \( \text{H}_2\text{O} \) it needs 15 oxygen atoms, fulfilled by 7.5 \( \text{O}_2 \) molecules.

Balancing the equation ensures that there are the same number of each type of atom on both sides of the chemical equation:\[2\text{C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2\text{O}\]This systematic balancing approach is essential in chemical reactions.

Moles and Molar Mass

The mole is a fundamental concept in chemistry used to measure the amount of substance. One mole contains Avogadro's number of entities (about \(6.022 \times 10^{23}\) particles). Molar mass is the mass of one mole of a substance, usually expressed in grams per mole.
For benzene (\(\text{C}_6\text{H}_6\)),
it has a molar mass of approximately 78.11 g/mol. Calculating moles helps quantify substances used and produced in a reaction.

• To find moles of benzene burned: Divide the mass of benzene (16.04 g) by its molar mass (78.11 g/mol).
  Result: \( \frac{16.04}{78.11} \approx 0.205 \text{ mol} \)
• This mole calculation allows determining how much oxygen and products are involved in the reaction. It's a cornerstone for further stoichiometry.

Understanding moles and molar mass is crucial in relating mass to quantities in a chemical equation.

Law of Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry stating that mass cannot be created or destroyed in a closed system through chemical reactions. This means that the mass of reactants is always equal to the mass of products in a chemical reaction.
During the combustion of benzene, we apply this law as follows:

• Mass of benzene burned: 16.04 g.
• Oxygen mass needed: 49.22 g (derived through stoichiometry using oxygen's molar mass).
• Total mass of products: Sum of reactants' mass (16.04 g + 49.22 g = 65.26 g).

The equality of reactant and product masses showcases the law at work. This principle is crucial in ensuring that chemical equations align with real-world scenarios, maintaining the matter's balance in any reaction.