/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 Assume that a single cylinder of... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 109

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3} .\) (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and 0.980 atm, how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

Short Answer

Expert verified
(a) First, we'll calculate the moles of air in the cylinder using the Ideal Gas Law: \(n = \frac{PV}{RT}\). Then, we'll find the moles of \(\mathrm{O}_{2}\) by multiplying the moles of air by the given mole fraction of \(\mathrm{O}_{2}\): \(Moles\,of\,O2 = Moles\,of\,air \cdot 0.2095\). (b) We'll use the balanced equation \(C_8H_{18} + 12.5O_2 \longrightarrow 8CO_2 + 9H_2O\) to find the moles of \(C_8H_{18}\) that can be combusted: \(Moles\,of\,C_8H_{18} = \frac{Moles\,of\,O2}{12.5}\). Finally, we'll convert the moles of \(C_8H_{18}\) to grams using the molar mass: \(Grams\,of\,C_8H_{18} = Moles\,of\, C_8H_{18} \cdot 114.22\,g/mol\).

Step by step solution

01

Calculate the moles of air present in the cylinder using the Ideal Gas Law

The Ideal Gas Law can be written as: \(PV=nRT\) We are given the temperature, pressure, and volume of the cylinder. We can use these values to find the moles (n) of air present in the cylinder. First, convert the temperature from Celsius to Kelvin. \(Temperature = 74^{\circ}\mathrm{C} + 273.15 = 347.15\mathrm{K}\) Now, convert the pressure to atmospheres: \(Pressure = 0.980\mathrm{atm}\) The volume of the cylinder is already given in cm鲁, so the values are in the desired units. Now, plug in the values and the gas constant \(R = 0.0821\frac{L \cdot atm}{mol \cdot K}\) to solve for the moles of air in the cylinder: \(n = \frac{PV}{RT}\)
02

Calculate the moles of O2 in the cylinder

To calculate the moles of O2 in the cylinder, use the mole fraction of O2 in the air, which is given as 0.2095: \(Moles\,of\,O2 = Moles\,of\,air \cdot Mole\,fraction\,of\,O2\) Now that we have the moles of O2 in the cylinder, we can proceed to find how many grams of C8H18 can be combusted by this quantity of O2. (b) Finding the grams of C8H18 that could be combusted by this quantity of O2
03

Write the balanced chemical equation for the combustion reaction

The combustion reaction of C8H18 is: \(C_8H_{18} + O_2 \longrightarrow CO_2 + H_2O\) Balancing this equation, we get: \(C_8H_{18} + 12.5O_2 \longrightarrow 8CO_2 + 9H_2O\)
04

Calculate the moles of C8H18 that can be combusted

Using the stoichiometry from the balanced equation, we can determine the moles of C8H18 that can be combusted: \(Moles\,of\,C_8H_{18} = \frac{Moles\,of\,O2}{12.5}\)
05

Convert the moles of C8H18 to grams

Lastly, using the molar mass of C8H18, we can convert the moles of C8H18 to grams: \(Molar\,mass\,of\,C_8H_{18} = 8\cdot(12.01\,g/mol) + 18\cdot(1.01\,g/mol) = 114.22\,g/mol\) \(Grams\,of\,C_8H_{18} = Moles\,of\, C_8H_{18} \cdot Molar\,mass\,of\,C_8H_{18}\) Now we have determined the grams of C8H18 that can be combusted by the quantity of O2 available in the cylinder.

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Key Concepts

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

Mole Fraction: Understanding Its Role
In chemistry, the mole fraction is a way of expressing the concentration of a component in a mixture. The mole fraction of a gas in a mixture is the ratio of the number of moles of that specific gas to the total number of moles of all gases in the mixture. This quantity is dimensionless and ranges from 0 to 1. In our exercise, we know that the mole fraction of oxygen, \(O_2\), in dry air is 0.2095. This means oxygen makes up approximately 20.95% of the air by mole count. This is crucial because it helps determine how many moles of oxygen are present in a given volume of air. To find the moles of \(O_2\) in our cylinder, we multiply the total moles of air calculated using the Ideal Gas Law by the mole fraction of \(O_2\). Whether you're looking to solve chemistry problems or understand real-world applications like engine performance, grasping the concept of mole fraction is essential.
Complete Combustion Explained
Complete combustion refers to a chemical reaction where a hydrocarbon, like octane (\(C_8H_{18}\)), reacts with oxygen (\(O_2\)) to produce carbon dioxide (\(CO_2\)) and water (\(H_2O\)). This kind of combustion is "complete" because it uses up all the hydrocarbon and oxygen in the reaction, leaving no unburned fuel or oxygen. The balanced chemical equation for the complete combustion of octane is necessary because it provides us the exact stoichiometric ratios. For every 1 mole of \(C_8H_{18}\) burned, 12.5 moles of \(O_2\) are needed. This ratio comes directly from balancing the chemical equation: \[C_8H_{18} + 12.5O_2 \rightarrow 8CO_2 + 9H_2O\]. Knowing this ratio helps us calculate how much fuel can be burned by the available oxygen. Complete combustion ensures maximum energy output from fuel, vital for efficiency in engines and other applications.
Stoichiometry: The Key to Calculations
Stoichiometry is the section of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It utilizes the coefficients from balanced chemical equations to determine how much of each substance is involved. In our exercise, once we know the moles of \(O_2\) in the cylinder, stoichiometry helps us find out how much \(C_8H_{18}\) (octane) can be combusted. Using the balanced combustion equation of octane, we see that 12.5 moles of \(O_2\) are needed for every mole of \(C_8H_{18}\). By dividing the moles of \(O_2\) by 12.5, we get the moles of \(C_8H_{18}\) that can react. Then, stoichiometry assists us in converting these moles into grams using the molar mass of octane. This process showcases how stoichiometry enables precise calculations in chemistry, crucial in fields such as fuel production and environmental science.

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