Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) burns in air: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ Balance the equation and determine the volume of air in liters at \(35.0^{\circ} \mathrm{C}\) and \(790 \mathrm{mmHg}\) required to burn \(227 \mathrm{~g}\) of ethanol. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume.

Short Answer

Expert verified
About 1850 liters of air are required to fully combust 227g of ethanol at 35.0°C and 790mmHg, given that air is 21.0% O2 by volume.

Step by step solution

01

Balance the Chemical Equation

In order to balance the equation, first count the number of atoms on both sides of the equation. You will see that there are 2 carbon atoms, 6 hydrogen atoms, and 1+2=3 oxygen atoms on the left side, and on the right side there are 1 carbon atom, 2 hydrogen atoms, and 2+1=3 oxygen atoms. To balance, add coefficients: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(l)\)
02

Calculate Number of Moles of Ethanol

Molar mass of ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) is about 46.07 g/mol. Using the given mass of 227g: Number of moles of ethanol = mass/molar mass = 227g / 46.07g/mol = approximately 4.92 moles.
03

Determine Moles of Oxygen Needed

From the balanced equation, 1 mol of ethanol requires 3 moles of oxygen. So, 4.92 moles of ethanol will require 4.92 * 3 = 14.76 moles of oxygen.
04

Calculate Volume of Air Needed

Assuming air is 21.0% O2 by volume, to find the volume of air required, simply divide the moles of oxygen needed by the volume percent: 14.76 moles / 0.21 (21% as a decimal) = approximately 70.29 moles worth of air.
05

Use the Ideal Gas Law

To convert from moles to liters, use the ideal gas law. The conditions for the gas are given: T=35.0°C=308.15K (since you must always convert to Kelvin in gas problems), P=790mmHg≈1.039atm (since 1 atm =760 mmHg), and R=0.0821L.atm/mol.K (commonly used value). The ideal gas law equation, PV=nRT, becomes V=nRT/P. So, plug in the numbers: Volume = (70.29 mol)(0.0821 L.atm/mol.K)(308.15 K) / 1.039 atm = approximately 1850L

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is the sect of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. For students grappling with balancing chemical equations, it's essential to understand that stoichiometry is not just about balancing atoms on both sides of the equation, but it's also about understanding the ratios in which substances react and form products.

Consider the burning of ethanol in the presence of oxygen to produce carbon dioxide and water. Here, the stoichiometry shows a one-to-three-to-two-to-three molar ratio, respectively. This ratio means that for each mole of ethanol, you'll need three moles of oxygen to ensure the complete conversion into two moles of carbon dioxide and three moles of water.

By understanding stoichiometry, you can comprehend the importance of reactant amounts, predict product yields, and calculate reactants or products needed for desired chemical outcomes. This background is especially valuable when scaling reactions for lab work or industrial processes.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas, through the equation PV=nRT. Here, R is the ideal gas constant which has several values depending on the unit system in use.

The ideal gas law allows us to interpret the behavior of gases under different conditions. In our problem with the combustion of ethanol, we determine the volume of air required using this law, taking into account the temperature and pressure conditions given. It's crucial to convert temperature to Kelvin and pressure to atmospheres to use the ideal gas law correctly.

Learning to apply the ideal gas law also serves as a foundation for understanding more complex gas behaviors and can guide you through other topics such as gas stoichiometry, partial pressures, and kinetic molecular theory.
Molar Mass
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It's directly equivalent to the substance's atomic or molecular weight taken from the periodic table. In the case of ethanol (\( \text{C}_{2}\text{H}_{5}\text{OH} \)), you need to calculate the molar mass by summing the atomic weights of each atom involved in its formula.

The molar mass serves as a bridge between the mass of a substance and the number of moles, which is paramount in stoichiometry to convert from grams to moles and vice versa. For any chemical reaction, calculating the molar mass is the initial step towards uncovering the stoichiometric relationships that define the reactants’ conversion into products. A deep familiarity with molar mass directly translates into the ability to successfully manage various quantitative aspects in chemistry experiments and real-world applications.

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Most popular questions from this chapter

The gas laws are vitally important to scuba divers. The pressure exerted by \(33 \mathrm{ft}\) of seawater is equivalent to 1 atm pressure. (a) A diver ascends quickly to the surface of the water from a depth of \(36 \mathrm{ft}\) without exhaling gas from his lungs. By what factor will the volume of his lungs increase by the time he reaches the surface? Assume that the temperature is constant. (b) The partial pressure of oxygen in air is about 0.20 atm. (Air is 20 percent oxygen by volume.) In deep-sea diving, the composition of air the diver breathes must be changed to maintain this partial pressure. What must the oxygen content (in percent by volume) be when the total pressure exerted on the diver is 4.0 atm? (At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gases.) (Hint: See the Chemistry in Action essay "Scuba Diving and the Gas Laws" in Section \(5.6 . ?\)

A sample of zinc metal reacts completely with an excess of hydrochloric acid: $$ \operatorname{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ The hydrogen gas produced is collected over water at \(25.0^{\circ} \mathrm{C}\) using an arrangement similar to that shown in Figure \(5.15 .\) The volume of the gas is \(7.80 \mathrm{~L},\) and the pressure is \(0.980 \mathrm{~atm} .\) Calculate the amount of zinc metal in grams consumed in the reaction. (Vapor pressure of water at \(25^{\circ} \mathrm{C}=23.8 \mathrm{mmHg} .\) )

A piece of sodium metal reacts completely with water as follows: $$ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) $$ The hydrogen gas generated is collected over water at \(25.0^{\circ} \mathrm{C}\). The volume of the gas is \(246 \mathrm{~mL}\) measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at \(25^{\circ} \mathrm{C}=0.0313\) atm. \()\)

What are the basic assumptions of the kinetic molecular theory of gases? How does the kinetic molecular theory explain Boyle's law, Charles' law, Avogadro's law, and Dalton's law of partial pressures?

The volume of a sample of pure HCl gas was \(189 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\) and \(108 \mathrm{mmHg} .\) It was completely dissolved in about \(60 \mathrm{~mL}\) of water and titrated with an \(\mathrm{NaOH}\) solution; \(15.7 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution were required to neutralize the \(\mathrm{HCl}\). Calculate the molarity of the \(\mathrm{NaOH}\) solution.

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