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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 \(99.3 \mathrm{kPa}\), 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
The given cylinder has 0.00438 moles of O鈧 present. About 0.04 grams of C鈧圚鈧佲倛 can be combusted by this quantity of O鈧, assuming complete combustion with the formation of CO鈧 and H鈧侽.

Step by step solution

01

Calculate the number of moles of air in the cylinder.

We will use the Ideal Gas Law to find the total number of moles of air in the cylinder: \(PV = nRT\) Where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the gas constant (0.0821 atm L/mol K), and T is the temperature (in Kelvin). First, we need to convert the given values: - Volume (1 cm鲁 = 0.001 L): 524 cm鲁 = 0.524 L - Temperature (Kelvin = Celsius + 273.15): 74掳C = 347.15 K - Pressure (1 kPa = 0.009869 atm): 99.3 kPa = 0.98644 atm Now, rearrange the ideal gas law equation to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and calculate n: \(n = \frac{(0.98644)(0.524)}{(0.0821)(347.15)} = 0.0209 \, \mathrm{moles \, of \, air}\)
02

Calculate the number of moles of O鈧 in the cylinder.

Now, we will use the mole fraction of O鈧 in the air to find the number of moles of O鈧 in the cylinder: \[n_{O_2} = n_{\mathrm{air}} \times X_{O_2}\] Where n鈧愥耽岬 is the number of moles of air and X_O鈧 is the mole fraction of O鈧. Plug in the values and calculate the number of moles of O鈧: \(n_{O_2}= (0.0209)(0.2095)= 0.00438\, \mathrm{moles\, of\, O_2}\) Now, we have the number of moles of O鈧 in the cylinder: \(0.00438 \, \text{moles}\).
03

Use the combustion reaction formula and find the grams of C鈧圚鈧佲倛 that can be combusted.

The balanced combustion reaction for C鈧圚鈧佲倛 with O鈧, forming CO鈧 and H鈧侽 is: \[ C_8H_{18} + 12.5O_2 \to 8CO_2 + 9H_2O\] From the balanced equation, we can see that 12.5 moles of O鈧 react with 1 mole of C鈧圚鈧佲倛. Now, set up the ratio to find the number of moles of C鈧圚鈧佲倛: \( \frac{n_{C_8H_{18}}}{n_{O_2}}=\frac{1}{12.5}\) Solve for n_C8H18: \(n_{C_8H_{18}} = \frac{1}{12.5} \times 0.00438=\ 0.000350\, \mathrm{moles\, of\, C_8H_{18}}\) Finally, calculate the mass of C鈧圚鈧佲倛 using its molar mass (C=12.01 g/mol, H=1.01 g/mol): \( M_{C_8H_{18}}=12.01 \times 8 + 1.01 \times 18 = 114.22\, \mathrm{g/mol}\) Now, calculate the grams of C鈧圚鈧佲倛: \(g_{C_8H_{18}}=0.000350\,\mathrm{moles\, of\, C_8H_{18}} \times 114.22\, \mathrm{g/mol}\) \(g_{C_8H_{18}} \, \approx \, 0.04 \, \mathrm{g}\) So, about 0.04 grams of C鈧圚鈧佲倛 can be combusted by the given quantity of O鈧 in the cylinder, assuming complete combustion with the formation of CO鈧 and H鈧侽.

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Key Concepts

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

Mole Fraction
The concept of mole fraction is a simple yet important one in chemistry. It is defined as the ratio of the moles of a component to the total moles of all components in a mixture. This dimensionless number expresses the concentration of a component in the mixture.

In dry air, which consists of several gases, oxygen ( O_2 ) is present with a specific mole fraction. Given that the mole fraction of oxygen in dry air is 0.2095, it means, proportionally, if air contains 1 mole of the gaseous mixture, approximately 0.2095 moles of it is O_2 .

To find the quantity of O_2 in a given volume, such as the cylinder in an engine, you first calculate the total moles of air using the Ideal Gas Law. Then, using the mole fraction, you can find how many moles of oxygen are actually present in that volume. This is crucial for understanding how much oxygen is available for reactions such as combustion.
Combustion Reaction
Combustion reactions are exothermic chemical processes where a fuel reacts with an oxidant, typically oxygen, resulting in the release of energy. The reaction often produces heat and light, and the most common products are CO_2 (carbon dioxide) and H_2O (water), especially in the complete combustion of hydrocarbons like C_8H_{18} (octane).

The balanced equation for the combustion of octane is:\[ C_8H_{18} + 12.5O_2 \to 8CO_2 + 9H_2O \]This equation tells us that 1 mole of octane reacts with 12.5 moles of oxygen. Understanding this stoichiometric relationship is critical because it allows you to calculate how much fuel can be combusted by a given quantity of oxygen. It lays the groundwork for concepts like fuel efficiency and pollutant control in engines.

In practical applications, knowing the amount of oxygen ensures the engine can achieve complete combustion, maximizing energy output and reducing unwanted byproducts from incomplete combustion.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). It is derived from the atomic masses of the elements that compose the compound, which are found on the periodic table.

For example, in the calculation of the molar mass of octane ( C_8H_{18} ), you sum the total mass of all carbon and hydrogen atoms in the molecule: - Carbon (C): 8 atoms 脳 12.01 g/mol = 96.08 g/mol - Hydrogen (H): 18 atoms 脳 1.01 g/mol = 18.18 g/mol Therefore, the molar mass of C_8H_{18} is the sum, 114.26 g/mol.

This fundamental calculation is essential for determining how much mass of a substance corresponds to a given number of moles, thus allowing you to convert moles to grams in reaction equations. It provides a bridge between the macroscopic properties of matter and the microscopic nature of atoms and molecules. Molar mass calculations are crucial for quantitative chemical analyses and must be conducted precisely to ensure accuracy in chemical computations.

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Most popular questions from this chapter

(a) Are you more likely to see the density of a gas reported in \(\mathrm{g} / \mathrm{mL}, \mathrm{g} / \mathrm{L},\) or \(\mathrm{kg} / \mathrm{cm}^{3} ?(\mathbf{b})\) Which units are appropriate for expressing atmospheric pressures, \(\mathrm{N}, \mathrm{Pa}, \mathrm{atm}, \mathrm{kg} / \mathrm{m}^{2} ?\) (c) Which is most likely to be a gas at room temperature and ordinary atmospheric pressure, \(\mathrm{F}_{2}, \mathrm{Br}_{2}, \mathrm{~K}_{2} \mathrm{O} .\)

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) 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 \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

A glass vessel fitted with a stopcock valve has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with \(\mathrm{Ar}\), it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

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