Question

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is responsible for the effects of intoxication felt after drinking alcoholic beverages. When ethanol burns in oxygen, carbon dioxide, and water are produced. (a) Write a balanced equation for the reaction. (b) How many liters of ethanol \(\left(d=0.789 \mathrm{~g} / \mathrm{cm}^{3}\right)\) will produce \(1.25 \mathrm{~L}\) of water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\) (c) A wine cooler contains \(4.5 \%\) ethanol by mass. Assuming that only the alcohol burns in oxygen, how many grams of wine cooler need to be burned to produce \(3.12 \mathrm{~L}\) of \(\mathrm{CO}_{2}\left(d=1.80 \mathrm{~g} / \mathrm{L}\right.\) at \(25^{\circ} \mathrm{C}, 1\) atm pressure) at the conditions given for the density?

Short answer

The balanced equation for the reaction of ethanol with oxygen is: $$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$$ b) How many cubic centimeters of ethanol are needed to produce 1.25 L of water? Approximately 1349.9 cm³ of ethanol are needed to produce 1.25 L of water. c) What mass of wine cooler needs to be burned to produce 3.12 L of CO2 under the given conditions? Approximately 64.9 g of wine cooler need to be burned to produce 3.12 L of CO2 under the given conditions.

Step-by-step solution

(a) Balanced equation for the reaction

To write the balanced equation for the reaction of ethanol and oxygen, we first need to identify the products formed, which are given as carbon dioxide (CO2) and water (H2O). The unbalanced equation is: $$C_2H_5OH + O_2 \rightarrow CO_2 + H_2O$$ Now, we need to balance the equation by adjusting the coefficients: $$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$$

(b) Volume of ethanol to produce 1.25 L of water

First, we need to find the number of moles of water formed, and then use the stoichiometry of the balanced equation to determine the number of moles of ethanol required. Finally, we will find the volume of ethanol. 1. Calculate the mass of water produced: \(1.25~L * 1~g/cm^3 * 1000~cm^3/L = 1250~g\) 2. Determine the number of moles of water produced: \(1250~g / 18.02~g/mol ≈ 69.37~mol\) 3. Use the stoichiometry from the balanced equation to find the number of moles of ethanol: \(69.37~mol_{H_2O} * (1~mol_{C_2H_5OH} / 3~mol_{H_2O}) ≈ 23.12~mol_{C_2H_5OH}\) 4. Determine the mass of ethanol: \(23.12~mol * 46.07~g/mol ≈ 1064.7~g\) 5. Calculate the volume of ethanol: \(1064.7~g / 0.789~g/cm^3 ≈ 1349.9~cm^3\) Therefore, approximately 1349.9 cm³ of ethanol are needed to produce 1.25 L of water.

(c) Mass of wine cooler to produce 3.12 L of CO2

To determine the mass of wine cooler needed to produce the desired amount of CO2, we will first find the number of moles of CO2 produced, then use stoichiometry to obtain the number of moles of ethanol needed, and finally determine the mass of wine cooler containing the required amount of ethanol. 1. Calculate the mass of CO2 produced: \(3.12~L * 1.80~g/L = 5.616~g\) 2. Determine the number of moles of CO2 produced: \(5.616~g / 44.01~g/mol ≈ 0.127~mol\) 3. Use the stoichiometry from the balanced equation to find the number of moles of ethanol: \(0.127~mol_{CO_2} * (1~mol_{C_2H_5OH} / 2~mol_{CO_2}) = 0.0635~mol_{C_2H_5OH}\) 4. Determine the mass of ethanol: \(0.0635~mol * 46.07~g/mol ≈ 2.92~g\) 5. Calculate the mass of wine cooler needed: \(2.92~g / 0.045 ≈ 64.9~g\) Therefore, approximately 64.9 g of wine cooler need to be burned to produce 3.12 L of CO2 under the given conditions.

Key concepts

Chemical Equation Balancing

Chemical equation balancing is crucial for understanding reactions like the combustion of ethanol. In this process, we ensure the number of each type of atom is equal on both sides of the equation. This maintains the conservation of mass principle.
When ethanol combusts, the primary reaction involves ethanol reacting with oxygen to form carbon dioxide and water. The initial unbalanced equation is: \[ \mathrm{C}_2\mathrm{H}_5\mathrm{OH} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]To balance it, we adjust coefficients to have the same number of carbon, hydrogen, and oxygen atoms on both sides. The balanced equation looks like this:\[ \mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O} \]Balanced chemical equations are foundational because they allow us to perform accurate stoichiometry calculations, which are necessary for determining the amounts of reactants and products.

Stoichiometry Calculations

Stoichiometry calculations rely on balanced chemical equations to determine how much of each substance is involved in a reaction. This involves using mole ratios from the balanced equation to convert between different substances.
For ethanol combustion, once we have the balanced equation, we can find out how moles of ethanol corresponds to moles of either oxygen, carbon dioxide, or water.

• The ratio of ethanol to carbon dioxide is 1:2.
• For water, the ratio is 1:3.

These ratios help us to determine the necessary amounts of reactants to produce a desired amount of a product, allowing for precise chemical preparation.
Stoichiometry is a powerful tool in chemistry, ensuring reactants are used efficiently and products are predicted accurately.

Gas Volume and Density

Understanding gas volume and density is essential when dealing with reactions involving gases. Volume relates to the space occupied, while density links mass to volume.
For a gas like carbon dioxide, the volume can determine its mass when the density is known. The formula to calculate volume given mass and density is:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]In ethanol combustion, knowing the density helps calculate how much ethanol is needed to produce a specific volume of CO₂ or H₂O. For example, to produce 1.25 L of water, we can use its density (1.00 g/cm³) to find the corresponding mass and, through stoichiometry, relate it back to ethanol consumed.Thus, mastering these concepts aids in predicting and understanding the behavior of gases during chemical reactions.

Moles and Molar Mass

The concepts of moles and molar mass are central to virtually all stoichiometry problems. A mole is a standard method in chemistry to express amounts of a chemical substance, based on Avogadro's number \(6.022 \times 10^{23}\) entities.
Molar mass, expressed in grams per mole, bridges moles to grams, allowing us to measure substances in a lab. For example, ethanol has a molar mass of 46.07 g/mol.
We use these values to convert between mass and moles. For 1250 g of water (molar mass of 18.02 g/mol), using the relation \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]we calculate the number of moles. These conversions are essential to determine how much reactant is needed to produce a specific amount of product.

Reaction Stoichiometry

Reaction stoichiometry puts all these concepts together to describe quantitatively the relationship between reactants and products in a chemical reaction.
With ethanol combustion, once we have a balanced equation, we utilize each term's stoichiometric coefficient to determine the necessary amount of reactants to achieve a desired amount of products. In the equation:\[ \mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O} \]For every mole of ethanol burned, two moles of carbon dioxide and three moles of water are produced. By using stoichiometric ratios, we ensure all molecules are adequately accounted for, ensuring no waste of materials.
Understanding reaction stoichiometry amplifies our capacity to predict the outcomes of chemical reactions and ensures efficiency in chemical syntheses.