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Chapter 5: Problem 162

A steel cylinder contains 5.00 mole of graphite (pure carbon) and 5.00 moles of \(\mathrm{O}_{2} .\) The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0\(\%\) . Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

Short Answer

Expert verified
The mole fractions of CO, CO\(_2\), and O\(_2\) after the combustion are approximately 0.291, 0.564, and 0.145, respectively.

Step by step solution

01

Write the balanced chemical equations

We have two possible reactions occurring in the cylinder: 1. Graphite (C) reacts with oxygen (O\(_2\)) to form carbon monoxide (CO): \(C + \frac{1}{2}O_2 \rightarrow CO\) 2. Graphite (C) reacts with oxygen (O\(_2\)) to form carbon dioxide (CO\(_2\)): \(C + O_2 \rightarrow CO_2\)
02

Set up the relationship between moles of CO and CO\(_2\)

Let x moles of CO be formed in the reaction, then we have 5 - x moles of CO\(_2\) formed. (5 moles of Carbon are used up)
03

Determine the total moles and moles of O\(_2\) left

Since we started with 5 moles of Oxygen, oxygen moles consumed to produce CO = \(\frac{1}{2}\) * x moles, and consumed to produce CO\(_2\) = (5 - x) moles. So, moles of O\(_2\) left = 5 - \(\frac{1}{2}\)x - 5 + x = \(\frac{1}{2}\)x moles. Total moles after reaction = x + 5 - x + \(\frac{1}{2}\)x = 5 + \(\frac{1}{2}\)x.
04

Calculate the change in pressure

We will use the ideal gas law equation, \(PV=nRT\). Since the temperature and the volume remain constant, the initial and final pressure ratios will relate to their respective mole ratios. The increase in pressure is 17%: \(\frac{P_{final}}{P_{initial}} = 1 + 0.17 = 1.17\) We can write this relation in terms of moles: \(\frac{5 + \frac{1}{2}x}{5} = 1.17\) (Because initial moles were 5)
05

Solve for x

Solve the equation from step 4 for x: \(\frac{1}{2}x = 5(0.17)\) x = 10(0.17) x = 1.7 moles
06

Calculate mole fractions

Now that we have the moles of CO (1.7), we can find the moles of CO\(_2\) formed as well (5 - 1.7 = 3.3). And we also find that \(\frac{1}{2}\) * 1.7 = 0.85 moles of O\(_2\) are left. Mole fraction of CO = \(\frac{1.7}{1.7 + 3.3 + 0.85} = \frac{1.7}{5.85}\) ≈ 0.291 Mole fraction of CO\(_2\) = \(\frac{3.3}{5.85}\) ≈ 0.564 Mole fraction of O\(_2\) = \(\frac{0.85}{5.85}\) ≈ 0.145 After the combustion reaction, the mole fractions of CO, CO\(_2\), and O\(_2\) in the final mixture are approximately 0.291, 0.564, and 0.145, respectively.

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Most popular questions from this chapter

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

At \(0^{\circ} \mathrm{C}\) a \(1.0-\mathrm{L}\) flask contains $5.0 \times 10^{-2}\( mole of \)\mathrm{N}_{2}, 1.5 \times\( \)10^{2} \mathrm{mg} \mathrm{O}_{2},\( and \)5.0 \times 10^{21}\( molecules of \)\mathrm{NH}_{3} .$ What is the partial pressure of each gas, and what is the total pressure in the flask?

An important process for the production of acrylonitrile $\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)$ is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A 150 -L reactor is charged to the following partial pressures at $25^{\circ} \mathrm{C} :$ $$\begin{aligned} P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\ P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\ P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa} \end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

Do all the molecules in a 1 -mole sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at 273 \(\mathrm{K}\)? Do all molecules in a 1 -mole sample of \(\mathrm{N}_{2}(g)\) have the same velocity at 546 \(\mathrm{K}\) ? Explain.

Calculate the pressure exerted by 0.5000 mole of \(\mathrm{N}_{2}\) in a 10.000 -L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 123.
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