Problem 25 - Chemistry and Chemical Reactivity

Question

Hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) burns in air \(\left(\mathrm{O}_{2}\right)\) to give \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write a balanced equation for the reaction. (b) If \(215 \mathrm{g}\) of \(\mathrm{C}_{6} \mathrm{H}_{14}\) is mixed with \(215 \mathrm{g}\) of \(\mathrm{O}_{2},\) what masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are produced in the reaction? (c) What mass of the excess reactant remains after the hexane has been burned?

Step-by-Step Solution

Answer

Hexane burns as \(215 \text{ g of } \mathrm{C}_{6}\mathrm{H}_{14}\) produces 186.56 g \(\mathrm{CO}_{2}\), 89.10 g \(\mathrm{H}_{2}\mathrm{O}\), with 153.94 g \(\mathrm{C}_{6}\mathrm{H}_{14}\) excess.

Step 1: Write the balanced chemical equation

To balance the combustion reaction of hexane, write the unbalanced equation: \(\mathrm{C}_{6}\mathrm{H}_{14} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\). Balance the carbon atoms with 6 \(\mathrm{CO}_{2}\), hydrogen atoms with 7 \(\mathrm{H}_{2}\mathrm{O}\), and oxygen atoms with 19/2 \(\mathrm{O}_{2}\) or equivalently 9.5 \(\mathrm{O}_{2}\), resulting in: \[2\mathrm{C}_{6}\mathrm{H}_{14} + 19\mathrm{O}_{2} \rightarrow 12\mathrm{CO}_{2} + 14\mathrm{H}_{2}\mathrm{O}\].

Step 2: Determine the molar masses

Calculate the molar mass of \(\mathrm{C}_{6}\mathrm{H}_{14}\) (6×12 + 14×1 = 86 \text{ g/mol}), and \(\mathrm{O}_{2}\) (2×16 = 32 \text{ g/mol}).

Step 3: Convert grams to moles

Convert the given masses to moles: - For \(\mathrm{C}_{6}\mathrm{H}_{14}\), \( \frac{215}{86} \approx 2.5 \text{ moles} \).- For \(\mathrm{O}_{2}\), \( \frac{215}{32} \approx 6.72 \text{ moles} \).

Step 4: Identify the limiting reactant

The balanced equation shows 2 moles of \(\mathrm{C}_{6}\mathrm{H}_{14}\) require 19 moles of \(\mathrm{O}_{2}\), meaning 2.5 moles \(\mathrm{C}_{6}\mathrm{H}_{14}\) require \(\frac{19}{2} \times 2.5 = 23.75 \) moles \(\mathrm{O}_{2}\). Since only 6.72 moles \(\mathrm{O}_{2}\) are available, \(\mathrm{O}_{2}\) is the limiting reactant.

Step 5: Calculate the moles of products

Given 6.72 moles of \(\mathrm{O}_{2}\) are used, calculate the products:- Use the ratio \( 19:12 \) from \(\mathrm{O}_{2}\) to \(\mathrm{CO}_{2}\): \( \frac{12}{19} \times 6.72 \approx 4.24 \text{ moles of } \mathrm{CO}_{2} \).- Use the ratio \( 19:14 \) from \(\mathrm{O}_{2}\) to \(\mathrm{H}_{2}\mathrm{O}\): \( \frac{14}{19} \times 6.72 \approx 4.95 \text{ moles of } \mathrm{H}_{2}\mathrm{O} \).

Step 6: Convert moles of products to grams

Calculate the masses:- \(\mathrm{CO}_{2}\) (molar mass = 44 g/mol): \( 4.24 \times 44 \approx 186.56 \text{ g} \).- \(\mathrm{H}_{2}\mathrm{O}\) (molar mass = 18 g/mol): \( 4.95 \times 18 \approx 89.10 \text{ g} \).

Step 7: Calculate mass of excess reactant

Determine moles of \(\mathrm{C}_{6}\mathrm{H}_{14}\) reacted using limit:- From the ratio \( 19:2 \), \( 6.72 \text{ moles of } \mathrm{O}_{2} \) corresponds to \( \frac{2}{19} \times 6.72 \approx 0.71 \text{ moles of } \mathrm{C}_{6}\mathrm{H}_{14} \).- Remaining \(\mathrm{C}_{6}\mathrm{H}_{14}\): \( 2.5 - 0.71 = 1.79 \text{ moles} \).- Convert to grams: \( 1.79 \times 86 \approx 153.94 \text{ g} \).

Key Concepts

Balancing Chemical Equations

In chemical reactions, balancing the equation is crucial to ensure that the law of conservation of mass is adhered to. This law states that mass cannot be created or destroyed in a closed system.
To balance a chemical equation, we must ensure the number of atoms for each element is the same on both sides of the equation.
For the combustion of hexane, we start with the unbalanced equation:

• \(\text{C}_6\text{H}_{14} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\)

1. Count the number of carbon atoms on the reactant side and match it by adding 6 \(\text{CO}_2\) on the product side.2. Count the hydrogen atoms on the reactant side and match it with 7 \(\text{H}_2\text{O}\) molecules.3. Balance the oxygen by adjusting the oxygen molecules \(\text{O}_2\) required, which balances as 19/2 or 9.5.
The balanced equation is:

• \(2\text{C}_6\text{H}_{14} + 19\text{O}_2 \rightarrow 12\text{CO}_2 + 14\text{H}_2\text{O}\)

This equation ensures that the number of each type of atom is equal on both sides of the reaction.

Limiting Reactant

The limiting reactant in a chemical reaction is the reactant that gets completely consumed first and limits the amount of products formed. When performing a reaction, identifying the limiting reactant is crucial as it determines the reaction's feasibility and the theoretical yield of products.
Using the stoichiometry of the balanced equation, we can find the limiting reactant:
Starting with the balanced equation, 2 moles of \(\text{C}_6\text{H}_{14}\) requires 19 moles \(\text{O}_2\). Given the quantities are 2.5 moles of \(\text{C}_6\text{H}_{14}\) and 6.72 moles of \(\text{O}_2\), we calculate:

• 2.5 moles of \(\text{C}_6\text{H}_{14}\) needs \(\frac{19}{2} \times 2.5 = 23.75\) moles \(\text{O}_2\).

However, only 6.72 moles \(\text{O}_2\) are available, so \(\text{O}_2\) is the limiting reactant.
It determines the maximum amount of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced.

Molar Mass

Molar mass is a critical concept in stoichiometry and involves the mass of one mole of a substance.
It allows us to convert between grams and moles, a necessary step in chemical calculations. For our reaction, finding the molar mass helps in converting reactant quantities from grams to moles. Let's consider:

• Calculate the molar mass of hexane \((\text{C}_6\text{H}_{14})\):
  • Carbon: \(12 \text{ g/mol} \times 6 = 72 \text{ g/mol}\)
  • Hydrogen: \(1 \text{ g/mol} \times 14 = 14 \text{ g/mol}\)
  Total = 86 g/mol.

For oxygen molecules \((\text{O}_2)\):

• \(2 \times 16 \text{ g/mol} = 32 \text{ g/mol}\).

These values enable the calculation of the number of moles of reactants and products, informing us how much of each substance is involved in the reaction.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the quantities of reactants and products in a chemical reaction. It relies on the mole ratio derived from the balanced equation, facilitating conversions between quantities of different substances.
In our example:

• We use the balanced equation \(2 \text{C}_6\text{H}_{14} + 19 \text{O}_2 \rightarrow 12 \text{CO}_2 + 14 \text{H}_2\text{O}\).

Each compound's coefficient in the equation represents a molar ratio. From this, we find that:

• \( \frac{12}{19} \) ratio from \(\text{O}_2\) to \(\text{CO}_2\), and \( \frac{14}{19} \) ratio from \(\text{O}_2\) to \(\text{H}_2\text{O}\) is used to calculate products:

Given 6.72 moles of \(\text{O}_2\), we find:

• \(\text{CO}_2\): \( \frac{12}{19} \times 6.72 \approx 4.24\) moles.
• \(\text{H}_2\text{O}\): \( \frac{14}{19} \times 6.72 \approx 4.95\) moles.

This stoichiometric analysis helps in predicting the yields of products based on available reactants.