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Propane is burned completely with excess oxygen. The product gas contains 24.5 mole \(\% \mathrm{CO}_{2}, 6.10 \%\) CO, \(40.8 \% \mathrm{H}_{2} \mathrm{O},\) and \(28.6 \% \mathrm{O}_{2}\) (a) Calculate the percentage excess \(\mathrm{O}_{2}\) fed to the furnace. (b) A student wrote the stoichiometric equation of the combustion of propane to form \(\mathrm{CO}_{2}\) and as $$2 \mathrm{C}_{3} \mathrm{H}_{8}+\frac{17}{2} \mathrm{O}_{2} \longrightarrow 3 \mathrm{CO}_{2}+3 \mathrm{CO}+8 \mathrm{H}_{2} \mathrm{O}$$ According to this equation, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) should be in a ratio of \(1 / 1\) in the reaction products, but in the product gas of Part (a) they are in a ratio of \(24.8 / 6.12 .\) Is that result possible? (Hint: Yes.) Explain how.

Short Answer

Expert verified
(a) The furnace was fed with about 70.05% excess oxygen. (b) The student's stoichiometric equation suggested equal amounts of CO2 and CO would be produced, but the actual ratio was about 4:1, indicating the production of CO2 was favored over CO in the combustion process.

Step by step solution

01

Calculate moles of products

The moles of different products given are percentages by mole, so let's assume we have a 100 moles of product gas to ease the calculations. We then have \(24.5\) moles of CO2, \(6.1\) moles of CO, \(40.8\) moles of H2O, and \(28.6\) moles of O2.
02

Calculate moles of reactants

The reaction for complete combustion of propane is given by: \(C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\). From the reaction, a mole of propane yields 3 moles of CO2 and 4 moles of H2O. Therefore, the numbers of moles of propane that reacted to form the given amounts of CO2 and H2O are: \[\frac{24.5}{3} = 8.1667 \text{ mol of } C_3H_8 \]\[\frac{40.8}{4} = 10.2 \text{ mol of } C_3H_8 \]We adopt the smaller value as the correct number of moles of propane that reacted, on the basis that any leftover amount would not have reacted due to lack of propane. Therefore number of moles of propane that reacted are \(8.1667\) moles.
03

Calculate excess O2

\(\mathrm{O}_{2}\) reacts with propane as per the reaction \(C_{3} H_{8}+5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} O\). Hence, for \(8.1667\) moles of \(C_{3} H_{8}\), the required moles of \(O_{2}\) are \(8.1667 \times 5 = 40.8335\) moles. However, in the combustion process, \(28.6\) moles of \(O_{2}\) were left unreacted in the product gas. Thus, the total moles of \(O_{2}\) fed into the furnace would have been \(40.8335 + 28.6 = 69.4335\) moles. The excess \(O_{2}\) as a percentage of the required amount is:\[\frac{\text{Excess amount}}{\text{Required amount}} \times 100 = \frac{28.6}{40.8335} \times 100 = 70.05 \%\]This means 70.05% excess oxygen was fed into the furnace.
04

Ratio Analysis

The student's stoichiometric equation implies that CO2 and CO should be in the ratio of 1:1. But, the actual ratio from the product gas is \(24.5/6.1= 4.01639:1\). In the actual reaction process, probably due to incomplete combustion, less CO was produced compared to the theoretical expectations. As combustion is a complex process, such discrepancies are possible even under controlled conditions.

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Key Concepts

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

Stoichiometry
Stoichiometry is at the heart of chemical reactions, serving as the 'mathematical recipe' for chemical processes. It relies on the law of conservation of mass, where matter is neither created nor destroyed during a reaction. This means that the reactants and products in a chemical equation must be equal in quantity of atoms for each element.

In the context of the combustion of propane, stoichiometry helps us deduce the precise amount of reactants needed to produce a set quantity of products. For each mole of propane (C_3H_8), stoichiometry tells us that we require 5 moles of oxygen (O_2) to yield 3 moles of carbon dioxide (CO_2) and 4 moles of water (H_2O). Understanding stoichiometry is vital not only for solving this problem, but also for a deeper grasp of how chemical reactions unfold.

In educational terms, visual aids such as 'mole maps' or 'stoichiometry road maps' can be highly beneficial for students. These visual tools assist in conceptualizing the stoichiometric relationships within a reaction, improving the ease with which one can perform calculations and predict the outcome of reactions.
Excess Reactant Calculation
Identifying an excess reactant is a key skill in chemistry, which prevents the waste of materials and ensures reaction efficiency. When calculating the excess reactant, we first determine the limiting reactant – the substance that will be completely consumed first in a chemical reaction, dictating the maximum amount of product produced. In our propane combustion example, by finding out that 8.1667 moles of propane were actually combusted, we can assert that propane is the limiting reactant.

With the limiting reactant identified, any other reactant present in greater quantities than necessary becomes an excess reactant. Knowing the stoichiometry of the reaction allows us to calculate the required amount of oxygen and compare it to the actual amount present. Subtracting the two gives us the excess quantity. Here, we learn that there was 70.05% excess oxygen.

To make this easier to follow, imagine if cooking a recipe called for 5 eggs per cake, but you had 10 eggs on hand. If you only bake one cake, you'd have 5 eggs left over - those 5 eggs represent an excess ingredient, similar to the excess oxygen in our chemical reaction.
Chemical Reaction Balancing
Balancing chemical reactions ensures that the law of conservation of mass is upheld. This entails making sure that the number of atoms for each element is equal on both sides of the reaction equation. For the combustion of propane, a balanced reaction is fundamental for accurately predicting the products and their amounts.

The initial confusion arises from the student's proposed stoichiometric equation, which incorrectly implies an equal ratio of CO_2 and CO production. A balanced combustion of propane would ideally produce only CO_2 and H_2O as products, but real-world phenomena like incomplete combustion can lead to the formation of CO as seen in the exercise.

Teaching this concept should emphasize the practice of balancing equations step by step, usually beginning with the most complex molecule. This trial-and-error skill strengthens students' understanding of molecular compositions and reaction mechanics. Additionally, introducing the concept of equilibrium and real-world reaction conditions can elucidate why reactions may not proceed to completion and why products like CO are present when theoretically, they shouldn't be.

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Most popular questions from this chapter

In the production of a bean oil, beans containing 13.0 wt\% oil and \(87.0 \%\) solids are ground and fed to a stirred tank (the extractor) along with a recycled stream of liquid \(n\) -hexane. The feed ratio is \(3 \mathrm{kg}\) hexane/kg beans. The ground beans are suspended in the liquid, and essentially all of the oil in the beans is extracted into the hexane. The extractor effluent passes to a filter where the solids are collected and form a filter cake. The filter cake contains 75.0 wt\% bean solids and the balance bean oil and hexane, the latter two in the same ratio in which they emerge from the extractor. The filter cake is discarded and the liquid filtrate is fed to a heated evaporator in which the hexane is vaporized and the oil remains as a liquid. The oil is stored in drums and shipped. The hexane vapor is subsequently cooled and condensed, and the liquid hexane condensate is recycled to the extractor. (a) Draw and label a flowchart of the process, do the degree-of-freedom analysis, and write in an efficient order the equations you would solve to determine all unknown stream variables, circling the variables for which you would solve. (b) Calculate the yield of bean oil product (kg oil/kg beans fed), the required fresh hexane feed \(\left(\mathrm{kg} \mathrm{C}_{6} \mathrm{H}_{14} / \mathrm{kg} \text { beans fed }\right),\) and the recycle to fresh feed ratio (kg hexane recycled/kg fresh feed). (c) It has been suggested that a heat exchanger might be added to the process. This process unit would consist of a bundle of parallel metal tubes contained in an outer shell. The liquid filtrate would pass from the filter through the inside of the tubes and then go on to the evaporator. The hot hexane vapor on its way from the evaporator to the extractor would flow through the shell, passing over the outside of the tubes and heating the filtrate. How might the inclusion of this unit lead to a reduction in the operating cost of the process? (d) Suggest additional steps that might improve the process economics.

Solid calcium fluoride (CaF, \(_{2}\) ) reacts with sulfuric acid to form solid calcium sulfate and gaseous hydrogen fluoride (HF). The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluoride is fluorite ore containing \(96.0 \mathrm{wt} \% \mathrm{CaF}_{2}\) and \(4.0 \% \mathrm{SiO}_{2}\) In a typical hydrofluoric acid manufacturing process, fluorite ore is reacted with 93 wt\% aqueous sulfuric acid, supplied 15\% in excess of the stoichiometric amount. Ninety-five percent of the ore dissolves in the acid. Some of the HF formed reacts with the dissolved silica in the reaction $$6 \mathrm{HF}+\mathrm{SiO}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ The hydrogen fluoride exiting from the reactor is subsequently dissolved in enough water to produce 60.0 wt\% hydrofluoric acid. Calculate the quantity of fluorite ore needed to produce a metric ton of aqueous hydrofluoric acid. Note: Some of the given data are not needed to solve the problem.

A stream consisting of 44.6 mole \(\%\) benzene and \(55.4 \%\) toluene is fed at a constant rate to a process unit that produces two product streams, one a vapor and the other a liquid. The vapor flow rate is initially zero and asymptotically approaches half of the molar flow rate of the feed stream. Throughout this entire period, no material accumulates in the unit. When the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole\% benzene. (a) Sketch a plot of liquid and vapor flow rates versus time from startup to when the flow rates become constant. (b) Is this process batch or continuous? Is it transient or steady-state before the vapor flow rate reaches its asymptotic limit? What about after it becomes constant? (c) For a feed rate of 100 mol/min, draw and fully label a flowchart for the process after the vapor flow rate has reached its limiting value, and then use balances to calculate the molar flow rate of the liquid and the composition of the vapor in mole fractions.

The popularity of orange juice, especially as a breakfast drink, makes this beverage an important factor in the economy of orange-growing regions. Most marketed juice is concentrated and frozen and then reconstituted before consumption, and some is "not-from-concentrate." Although concentrated juices are less popular in the United States than they were at one time, they still have a major segment of the market for orange juice. The approaches to concentrating orange juice include evaporation, freeze concentration, and reverse osmosis. Here we examine the evaporation process by focusing only on two constituents in the juice: solids and water. Fresh orange juice contains approximately 10 wt\% solids (sugar, citric acid, and other ingredients) and frozen concentrate contains approximately 42 wt\% solids. The frozen concentrate is obtained by evaporating water from the fresh juice to produce a mixture that is approximately 65 wt\% solids. However, so that the flavor of the concentrate will closely approximate that of fresh juice, the concentrate from the evaporator is blended with fresh orange juice (and other additives) to produce a final concentrate that is approximately 42 wt\% solids. (a) Draw and label a flowchart of this process, neglecting the vaporization of everything in the juice but water. First prove that the subsystem containing the point where the bypass stream splits off from the evaporator feed has one degree of freedom. (If you think it has zero degrees, try determining the unknown variables associated with this system.) Then perform the degree- offreedom analysis for the overall system, the evaporator, and the bypass- evaporator product mixing point, and write in order the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don't do any calculations. (b) Calculate the amount of product (42\% concentrate) produced per 100 kg fresh juice fed to the process and the fraction of the feed that bypasses the evaporator. (c) Most of the volatile ingredients that provide the taste of the concentrate are contained in the fresh juice that bypasses the evaporator. You could get more of these ingredients in the final product by evaporating to (say) 90\% solids instead of 65\%; you could then bypass a greater fraction of the fresh juice and thereby obtain an even better tasting product. Suggest possible drawbacks to this proposal.

n-Pentane is burned with excess air in a continuous combustion chamber. (a) A technician runs an analysis and reports that the product gas contains 0.270 mole\% pentane, \(5.3 \%\) oxygen, \(9.1 \%\) carbon dioxide, and the balance nitrogen on \(a\) dry basis. Assume 100 mol of dry product gas as a basis of calculation, draw and label a flowchart, perform a degree-offreedom analysis based on atomic species balances, and show that the system has -1 degree of freedom. Interpret this result. (b) Use balances to prove that the reported percentages could not possibly be correct. (c) The technician reruns the analysis and reports new values of 0.304 mole\% pentane, \(5.9 \%\) oxygen, \(10.2 \%\) carbon dioxide, and the balance nitrogen. Verify that this result could be correct and, assuming that it is, calculate the percent excess air fed to the reactor and the fractional conversion of pentane. (d) It was emphasized in Part (c) that the new composition could be correct. Explain why it isn't possible to say for sure; illustrate your response by considering a set of equations with -1 degree of freedom.

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