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Chapter 10: Problem 107

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 \(0.980 \mathrm{~atm}\), 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 \(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? as originally filled with gas?

Short Answer

Expert verified
When the cylinder is full at the given conditions, there are \(n_{O_2} = 0.0126\) moles of O鈧� present. This amount of O鈧� can combust approximately \(0.807 g\) of C鈧圚鈧佲倛, assuming complete combustion.

Step by step solution

01

Calculate the number of moles of air in the cylinder

We can find the number of moles of air in the cylinder by using the ideal gas law equation: \(PV = nRT\), where \(P =\) pressure, \(V =\) volume, \(n =\) number of moles, \(R =\) universal gas constant, and \(T =\) temperature. Given the volume \(V = 524 cm^3\), pressure \(P = 0.980 atm\), and temperature \(T = 74掳C\), we need to rewrite the temperature in Kelvin and convert the values to appropriate units before using the ideal gas equation. First, convert the temperature to Kelvin: \(T_K = T_C + 273.15\) \(T_K= 74 + 273.15\) \(T_K= 347.15 K\) Now, convert the volume from cm鲁 to L: \(V = 524 cm^3 * (1 L / 1000 cm^3) = 0.524 L\) The ideal gas equation becomes: \(P * V = n * R * T_K\) Plug in the given values, with \(R = 0.0821 L * atm / (mol * K)\), and solve for n: \(0.980 atm * 0.524 L = n * 0.0821 L * atm / (mol * K) * 347.15 K\)
02

Determine the number of moles of O鈧�

Given the mole fraction of O鈧� in dry air is 0.2095. Therefore, multiplying the number of total moles of air with the mole fraction of O鈧� will give us the number of moles of O鈧� present in the cylinder: \(n_{O_2} = n_{total} * 0.2095\)
03

Find the mass of C鈧圚鈧佲倛 that can be combusted using the quantity of O鈧�

Write the balanced chemical equation for the complete combustion of C鈧圚鈧佲倛 with O鈧�: \(C_8H_{18} + 12.5O_2 \rightarrow 8CO_2 + 9H_2O\) Use stoichiometry to find the number of moles of C鈧圚鈧佲倛 that can be combusted using the given amount of O鈧�: \(1 C_8H_{18} = 12.5 O_2\) Calculate the mass of C鈧圚鈧佲倛 that can be combusted by this amount of O鈧� using its molar mass (114.23 g/mol): \(mass_{C_8H_{18}} = n_{C_8H_{18}} * M_{C_8H_{18}}\) Replace M_{C_8H_{18}} with the molar mass of 114.23 g/mol and solve for mass_{C_8H_{18}}.

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Key Concepts

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

Moles of Gas
In chemistry, understanding the concept of moles is crucial as it forms the basis for the ideal gas law and other calculations. A mole is a standard scientific unit for measuring large quantities of very small entities, like atoms or molecules. When dealing with gases, the ideal gas law becomes a useful tool that correlates pressure, volume, temperature, and the number of moles using the equation: \[PV = nRT\]Here, \(P\) is the pressure, \(V\) is the volume, \(n\) represents the number of moles, \(R\) is the universal gas constant (0.0821 L路atm/mol路K), and \(T\) is the temperature in Kelvin. To calculate the number of moles of gas, ensure that all units are compatible. For example, convert volume into liters and temperature into Kelvin. In our given exercise, the conversion involved transforming 524 cm鲁 to 0.524 liters and 74掳C to 347.15 Kelvin. This conversion is crucial to apply the formula correctly. Once we plug these values into the equation, we can solve for \(n\), the moles of gas present.
Mole Fraction
The mole fraction is an important concept used in various branches of science, including chemistry. It represents the ratio of moles of a particular component to the total moles of all components in a mixture.For dry air, the mole fraction of \(\mathrm{O}_2\) is noted as 0.2095, meaning that out of the total moles of air, 20.95% are moles of oxygen. This proportion is crucial when calculating the specific amount of gas in a mixture. To find the moles of \(\mathrm{O}_2\) in a gas mixture, multiply the total moles of gas by the mole fraction of \(\mathrm{O}_2\). In the problem at hand, after determining the total moles of air using the ideal gas law, you multiply the total moles by 0.2095 to ascertain how many moles of \(\mathrm{O}_2\) are in the engine cylinder. This step is key in understanding the composition of air and performing subsequent combustion calculations.
Combustion Reaction
Combustion reactions are chemical processes where a fuel reacts with an oxidant (often \(O_2\)) to produce energy. In our exercise, the focus is on the complete combustion of octane (\(C_8H_{18}\)), a common automotive fuel.The balanced combustion equation for octane is:\[C_8H_{18} + 12.5O_2 \rightarrow 8CO_2 + 9H_2O\]This indicates that one mole of octane requires 12.5 moles of \(O_2\) to combust completely, forming carbon dioxide and water as byproducts. To find out how much octane can be burned using the moles of \(\mathrm{O}_2\) available, use stoichiometry.- Calculate moles of \(C_8H_{18}\) by dividing the moles of \(O_2\) by 12.5.- Multiply this quantity by the molar mass of \(C_8H_{18}\) (114.23 g/mol) to find the mass of octane that can be combusted.Understanding the stoichiometric relationships and balancing equations is vital for calculating reactants and products in chemical reactions, helping to closely predict how much fuel is needed for energy conversion.

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Most popular questions from this chapter

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is greater than the average speed of air molecules, and the greater speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The mass of the balloon is thus less than the mass of the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is greater than the air temperature. Hot gases tend to rise.

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is \(8.102 \mathrm{~g}\). (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\) ? (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL}\); the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

\(10.73\) A quantity of \(\mathrm{N}_{2}\) gas originally held at \(5.25 \mathrm{~atm}\) pressure in a \(1.00\) - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5\) - \(\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a \(5.00\)-L container is transferred to this same container. What is the total pressure in the new container?

Which assumptions are common to both kinetic-molecular theory and the ideal- gas equation?
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