Question

In 1997 and \(2009,\) at United Nations conferences on climate change, many nations agreed to expand their research efforts to develop renewable sources of carbon-based fuels. For more than a quarter century, Brazil has been engaged in a program to replace gasoline with ethanol derived from the root crop manioc (cassava). (a) Write separate balanced equations for the complete combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and of gasoline (represented by the formula \(\mathrm{C}_{8} \mathrm{H}_{18}\) ). (b) What mass (g) of oxygen is required to burn completely \(1.00 \mathrm{~L}\) of a mixture that is \(90.0 \%\) gasoline \((d=0.742 \mathrm{~g} / \mathrm{mL})\) and \(10.0 \%\) ethanol \((d=0.789 \mathrm{~g} / \mathrm{mL})\) by volume? (c) If \(1.00 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) occupies \(22.4 \mathrm{~L}\), what volume of \(\mathrm{O}_{2}\) is needed to burn \(1.00 \mathrm{~L}\) of the mixture? (d) Air is \(20.9 \% \mathrm{O}_{2}\) by volume. What volume of air is needed to burn \(1.00 \mathrm{~L}\) of the mixture?

Short answer

2502.848 g of \( \text{O}_2 \) is needed to burn the mixture. The volume of \( \text{O}_2 \) required is 1750.994 L. The volume of air required is 8377.58 L.

Step-by-step solution

- Balanced Equation for Ethanol Combustion

The chemical formula for ethanol is \(\text{C}_2\text{H}_5\text{OH}\). The complete combustion of ethanol can be represented by the reaction: \[\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O} \]

- Balanced Equation for Gasoline Combustion

The chemical formula for gasoline is typically represented as \(\text{C}_8\text{H}_{18}\). The complete combustion of gasoline can be represented by the reaction: \[ \text{C}_8\text{H}_{18} + 12.5\text{O}_2 \to 8\text{CO}_2 + 9\text{H}_2\text{O} \]

- Calculate Mass of Gasoline and Ethanol

Given the densities and volumes in the mixture, calculate the mass of each component. For gasoline: \[ \text{Mass of gasoline} = 0.90 \times 1000 \text{ mL} \times 0.742 \frac{\text{g}}{\text{mL}} = 667.8 \text{ g} \] For ethanol: \[ \text{Mass of ethanol} = 0.10 \times 1000 \text{ mL} \times 0.789 \frac{\text{g}}{\text{mL}} = 78.9 \text{ g} \]

- Moles of Gasoline and Ethanol

Convert the mass of gasoline and ethanol to moles. Molar masses: \( \text{Molar mass of gasoline} = 114.23 \frac{\text{g}}{\text{mol}} \) and \( \text{Molar mass of ethanol} = 46.07 \frac{\text{g}}{\text{mol}} \) For gasoline: \[ \text{Moles of gasoline} = \frac{667.8 \text{ g}}{114.23 \frac{\text{g}}{\text{mol}}} = 5.846 \text{ mol} \] For ethanol: \[ \text{Moles of ethanol} = \frac{78.9 \text{ g}}{46.07 \frac{\text{g}}{\text{mol}}} = 1.713 \text{ mol} \]

- Determine Oxygen Required

Using stoichiometry, calculate the moles of \( \text{O}_2 \) needed. For gasoline: \[ 5.846 \text{ mol of \text{C}_8\text{H}_{18}} \times 12.5 \text{ mol of \text{O}_2 / mol of \text{C}_8\text{H}_{18}} = 73.075 \text{ mol of \text{O}_2} \] For ethanol: \[ 1.713 \text{ mol of \text{C}_2\text{H}_5\text{OH}} \times 3 \text{ mol of \text{O}_2 / mol of \text{C}_2\text{H}_5\text{OH}} = 5.139 \text{ mol of \text{O}_2} \] Total moles of \( \text{O}_2 \) required: \[ 73.075 + 5.139 = 78.214 \text{ mol of \text{O}_2} \]

- Mass of Oxygen Needed

Calculate the total mass of \( \text{O}_2 \) needed: \[ \text{Mass of \text{O}_2} = 78.214 \text{ mol} \times 32 \frac{\text{g}}{\text{mol}} = 2502.848 \text{ g} \]

- Volume of \( \text{O}_2 \) Needed

Using the ideal gas law, calculate the volume of \( \text{O}_2 \) needed: \[ \text{Volume of \text{O}_2} = 78.214 \text{ mol} \times 22.4 \frac{\text{L}}{\text{mol}} = 1750.994 \text{ L} \]

- Volume of Air Needed

Given that air is 20.9% \( \text{O}_2 \) by volume, calculate the volume of air required: \[ \text{Volume of air} = \frac{1750.994 \text{ L}}{0.209} = 8377.58 \text{ L} \]

Key concepts

balanced chemical equations

To understand the combustion of ethanol and gasoline, first, we need to write balanced chemical equations. In a balanced equation, the number of atoms for each element is the same on both sides of the reaction.
For ethanol \(\text{C}_2\text{H}_5\text{OH}\): \(\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O}\).
For gasoline \( ( \text{C}_8\text{H}_{18}}\): \(\text{C}_8\text{H}_{18} + 12.5\text{O}_2 \to 8\text{CO}_2 + 9\text{H}_2\text{O}\).

These balanced equations show how ethanol and gasoline react with oxygen. They form carbon dioxide and water as products. Balancing these equations ensures all elements are conserved in the reaction, critical for accurate calculations involved in further steps.

stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. Here, it's used to determine the amount of oxygen needed for the combustion of a specific volume of ethanol-gasoline mixture.
For example, from the balanced equations, we can see that combusting one molecule of ethanol requires three molecules of oxygen. Similarly, one molecule of gasoline requires twelve and a half molecules of oxygen.
Knowing the densities and volumes of gasoline and ethanol in the mixture, we first calculate the mass of each component. Then, we convert these masses to moles using their respective molar masses.
Using stoichiometry, we then determine the moles of oxygen required for each component. Summing these values gives the total moles of oxygen required for the combustion of the mixture. This process illustrates the crucial role of stoichiometry in practical chemical calculations.

ideal gas law

The Ideal Gas Law equation is \( PV = nRT \), relating pressure (P), volume (V), amount of gas (n in moles), the gas constant (R), and temperature (T). For simplicity, in this problem, it's given that 1 mole of \(O_2\) occupies 22.4 liters at standard temperature and pressure (STP).
After calculating the total moles of oxygen required, we use this concept to find the volume of oxygen needed. For instance, if 78.214 moles of \(O_2\) are required, the corresponding volume is calculated as:
\[ \text{Volume of } O_2 = 78.214 \text{ mol} \times 22.4 \frac{\text{L}}{\text{mol}} = 1750.994 \text{ L} \].
This volume represents the oxygen needed for the complete combustion of the ethanol-gasoline mixture under standard conditions, illustrating the practical application of the ideal gas law in real-world scenarios.

oxygen requirements for combustion

Understanding the oxygen requirements is crucial for combustion analysis. For ethanol and gasoline, the balanced chemical equations already show how much oxygen each combustion needs.
Ethanol: \(\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O}\)
Gasoline: \(\text{C}_8\text{H}_{18} + 12.5\text{O}_2 \to 8\text{CO}_2 + 9\text{H}_2\text{O}\)
Combining these with stoichiometric calculations, we find the moles of \(O_2\) required for the given masses of fuel. By determining the total mass and volume of \(O_2\) needed, we assess how much oxygen is required for efficient combustion. This ensures complete fuel utilization, minimizing unburnt residues and optimizing the combustion process.

air composition

Air is composed of various gases, with oxygen making up approximately 20.9%. This composition is essential when calculating the volume of air needed for combustion.
Given the volume of oxygen needed, we use the percentage of oxygen in air to find the total air volume. For example, if 1750.994 liters of \(O_2\) are required: \[ \text{Volume of air} = \frac{1750.994 \text{ L }}{0.209} = 8377.58 \text{ L} \].
This tells us the air volume needed to provide the necessary oxygen. Understanding air composition helps us calculate and rationalize the fuel-to-air ratio required for complete combustion, highlighting the interplay between chemistry and environmental sciences.