Problem 69 - Chemistry Matter and Change

Question

Gasohol is a mixture of ethanol and gasoline. Balance the equation, and determine the mass of \(\mathrm{CO}_{2}\) produced from the combustion of 100.0 g of ethanol. \begin{equation} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \end{equation}

Step-by-Step Solution

Answer

Approximately 191.07 g of \(\mathrm{CO}_{2}\) is produced.

Step 1: Balance the Chemical Equation

Before solving the problem, we must ensure the chemical equation for combustion is balanced. The unbalanced equation is:\[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\mathrm{l}) + \mathrm{O}_{2} (\mathrm{g}) \rightarrow \mathrm{CO}_{2} (\mathrm{g}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{g})\]Balancing it, we get:\[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\mathrm{l}) + 3 \mathrm{O}_{2} (\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2} (\mathrm{g}) + 3 \mathrm{H}_{2} \mathrm{O} (\mathrm{g})\]This balanced equation indicates that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water.

Step 2: Determine the Molar Mass of Ethanol and CO2

The molecular formula for ethanol is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). Calculating its molar mass:\[2 \times 12.01 \,(\mathrm{C}) + 6 \times 1.008 \,(\mathrm{H}) + 16.00 \,(\mathrm{O}) = 46.08 \, \text{g/mol}\]The molar mass of \(\mathrm{CO}_{2}\) is calculated as:\[1 \times 12.01 \,(\mathrm{C}) + 2 \times 16.00 \,(\mathrm{O}) = 44.01 \, \text{g/mol}\]These values will be used to calculate how much \(\mathrm{CO}_{2}\) is produced.

Step 3: Calculate Moles of Ethanol

Using the mass provided, convert the mass of ethanol to moles. We have 100.0 g of ethanol.\[\text{moles of ethanol} = \frac{100.0 \, \text{g}}{46.08 \, \text{g/mol}} \approx 2.17 \, \text{mol}\]

Step 4: Use Stoichiometry to Find Moles of CO2

According to the balanced equation, 1 mole of ethanol produces 2 moles of \(\mathrm{CO}_{2}\). Therefore, calculate the moles of \(\mathrm{CO}_{2}\) produced from 2.17 moles of ethanol.\[2.17 \, \text{mol C}_2\mathrm{H}_5\mathrm{OH} \times \frac{2 \, \text{mol CO}_2}{1 \, \text{mol C}_2\mathrm{H}_5\mathrm{OH}} = 4.34 \, \text{mol CO}_2\]

Step 5: Calculate Mass of CO2 Produced

Finally, convert moles of \(\mathrm{CO}_{2}\) to mass using its molar mass.\[\text{mass of CO}_2 = 4.34 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 191.07 \, \text{g}\]Thus, 191.07 g of \(\mathrm{CO}_{2}\) is produced from the combustion of 100.0 g of ethanol.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a vital skill in stoichiometry. It ensures that the law of conservation of mass is adhered to, meaning that atoms are neither created nor destroyed in chemical reactions. In a balanced equation, the number of each type of atom on the reactant side must equal the number on the product side.
To balance the combustion of ethanol, begin with the unbalanced equation: \[\mathrm{C}_2\mathrm{H}_5 \mathrm{OH} (\mathrm{l}) + \mathrm{O}_2 (\mathrm{g}) \rightarrow \mathrm{CO}_2 (\mathrm{g}) + \mathrm{H}_2 \mathrm{O} (\mathrm{g})\]Start by balancing carbon atoms. Ethanol contains two carbon atoms, so two molecules of \(\mathrm{CO}_2\) should be produced. Next, balance hydrogen atoms; ethanol has six hydrogen atoms, thus forming three water molecules. Finally, balance the oxygen atoms last. Ethanol and two moles of \(\mathrm{CO}_2\) and three moles of \(\mathrm{H}_2 \mathrm{O}\) collectively require three moles of oxygen gas, achieving balance.
This gives us: \[\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) + 3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{g}) + 3 \mathrm{H}_2\mathrm{O}(\mathrm{g})\]
This equation is essential for further calculations.

Ethanol Combustion

Ethanol combustion is a common reaction involving ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and oxygen. It is often used in engines and laboratories because of its effectiveness in producing energy. In this reaction, ethanol combines with oxygen to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).
The reaction is exothermic, meaning it releases heat. The process provides insight into how ethanol, as an alcohol fuel, can serve as an energy source. Understanding ethanol combustion involves knowing both the inputs (ethanol and oxygen) and outputs (carbon dioxide and water), along with the energy changes involved.
By knowing the balanced equation from the previous section, the stoichiometry of the reaction can be understood, aiding in calculating how much product is formed from a given amount of reactant.

Molar Mass Calculation

Calculating molar mass is essential when dealing with stoichiometry. It’s the mass of one mole of a substance, usually in grams per mole (g/mol). For ethanol (\(\text{C}_2\text{H}_5\text{OH}\)), the molar mass calculation considers the atomic masses of carbon, hydrogen, and oxygen.
Each carbon atom weighs 12.01 g/mol, hydrogen is 1.008 g/mol, and oxygen is 16.00 g/mol.

• Ethanol Calculation: \(2 \times 12.01 + 6 \times 1.008 + 16.00 = 46.08\,\text{g/mol}\)
• Carbon Dioxide Calculation: Calculating the molar mass of \(\text{CO}_2\) involves adding the mass of one carbon atom and two oxygen atoms:\(12.01 + 2 \times 16.00 = 44.01\,\text{g/mol}\).

Understanding these calculations helps us convert between mass and moles in chemical equations—serving as a bridge between lab measurements and chemical reactions. Knowing these values is crucial for subsequent stoichiometry calculations.

Mass-to-Moles Conversion

Converting mass to moles is a crucial step in stoichiometry because it allows us to use the balanced chemical equation to predict the amount of products formed. The formula used to convert mass to moles is:\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given 100.0 g of ethanol, we use its molar mass (46.08 g/mol) for conversion. This calculation shows how many moles of ethanol are involved:\[ \frac{100.0 \text{ g}}{46.08 \text{ g/mol}} \approx 2.17 \text{ mol} \].
With the number of moles known, the balanced equation can be used to relate reactants to products. Since each mole of ethanol produces two moles of \(\mathrm{CO}_2\), we can then determine the moles of carbon dioxide produced. Calculating moles is a foundational part of translating the balanced chemical equation into meaningful, actionable data.