/*! 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 95 A mixture of propane and butane ... [FREE SOLUTION] | 91Ó°ÊÓ

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A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole \(\% \mathrm{H}_{2} \mathrm{O}\). After all the water is removed from the products, the residual gas contains 69.4 mole \(\% \mathrm{CO}_{2}\) and the balance \(\mathrm{O}_{2}\) (a) What is the mole percent of propane in the fuel? (b) It now turns out that the fuel mixture may contain not only propane and butane but also other hydrocarbons. All that is certain is that there is no oxygen in the fuel. Use atomic balances to calculate the elemental molar composition of the fuel from the given combustion product analysis (i.e., what mole percent is \(C\) and what percent is \(\mathrm{H}\) ). Prove that your solution is consistent with the result of Part (a).

Short Answer

Expert verified
The mole percent of propane in the fuel is 57.6%. The mole percent of Carbon and Hydrogen in the fuel is 66.9% and 33.1% respectively.

Step by step solution

01

Write the combustion reactions for propane and butane

For propane: \(C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\)\For butane: \(C_4H_{10} + 6.5O_2 \rightarrow 4CO_2 + 5H_2O\)
02

Analyze the mole percentages of the combustion products

From the problem, we know the combustion product contains 47.4 mole% \(H_2O\) and 69.4 mole% \(CO_2\) which makes up 116.8 mole % in total. We know that the remaining mole % is \(O_2\), which is \(100 - 116.8 = 16.2\% O_2\). Assuming total 100 moles of gasses are produced after combustion, there are 47.4 moles of \(H_2O\), 69.4 moles of \(CO_2\), and 16.2 moles of \(O_2\).
03

Calculate the mol percent of Propane in the fuel

From step 1, we can assume the fuel is composed of x% propane and (100 - x)% butane. For \(H_2O\); from propane combustion, x moles contribute 4x moles of \(H_2O\), and from butane combustion (100 - x) moles contribute 5*(100 - x) moles of \(H_2O\). From the equation 4x + 5*(100 - x) = 47.4, we compute x = 57.6%, thus the mole percent of Propane in the fuel is 57.6%.
04

Calculate the atomic balances to find the elemental molar composition

Given that there is no oxygen in the fuel, it can be assumed to be composed of C and H. Since the moles of \(H_2O\) equal to half the number of hydrogen atoms and the moles of \(CO_2\) represent the number of carbon atoms, the component composition of the fuel can be calculated as: Carbon moles % = \(69.4/ (69.4 + 47.4/2)*100 = 66.9 % \) and Hydrogen mole% = \(100 - 66.9 = 33.1 %\). Therefore, the mole percent of C is 66.9 and H is 33.1.
05

Check if the found solution is consistent the part (a) result

Calculating the composition of butane and propane we get: Propane C = (57.6% of propane)*3C = 1.8C and H = (57.6% of propane)*8H = 4.61H, Butane C = (100-57.6)% of butane)*4C = 1.68C and H = (100-57.6)% of butane)*10H = 4.21H. When these values are added, they give a C content of 1.8+1.68 = 3.48 and an H content of 4.61+4.21 = 8.82. Making a direct comparison to the moles of C and H found in Step 4 indicates consistency.

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Key Concepts

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

Combustion Analysis
Combustion analysis is a method used to understand the characteristics and components of a burning process. In our scenario, a mixture of propane and butane, both of which are hydrocarbons, is combusted with pure oxygen. The products of this combustion process are primarily carbon dioxide \( (CO_2) \) and water \( (H_2O) \).
To analyze the results of this chemical reaction, we observe the mole percentages of the combustion products. By knowing the quantities of \( CO_2 \) and \( H_2O \), we can deduce the composition of the original fuel mixture. Essentially, combustion analysis helps us identify and quantify the substances present in the initial fuel based on the gaseous products formed.
In practice, this involves balancing chemical equations, analyzing atomic balances, and interpreting the results. It allows us to reverse engineer the combustion process to understand the fuel composition in terms of the chemical elements such as carbon and hydrogen. This method is widely used in chemical reaction engineering to determine specific details about fuel sources and combustion efficiency.
Mole Percent Calculation
Mole percent calculation is a vital concept in chemical reaction engineering. It allows us to determine the composition of a mixture in terms of the different components present. In this context, we are dealing with a combustion problem where various products are formed.
To find the mole percent of a component in a mixture, you calculate the number of moles of that component, divide it by the total number of moles of all substances involved, and multiply by 100 to get a percentage.
In our example, after combustion, we calculated the mole percentages of \( H_2O \) and \( CO_2 \) from the reaction. By further breaking down the results using the balanced chemical equations, we can obtain the mole percentages of propane and butane in the original fuel. Understanding how mole percent calculations work is crucial for analyzing combustion reactions and other chemical processes.
Hydrocarbon Fuels
Hydrocarbon fuels, such as propane and butane, are organic chemical compounds composed primarily of carbon and hydrogen atoms. These fuels are commonly used because they efficiently release energy when combusted. Propane \( (C_3H_8) \) and butane \( (C_4H_{10}) \) are both part of the alkane family and are gaseous at room temperature.
When these hydrocarbons combust in the presence of oxygen, they produce carbon dioxide \( (CO_2) \) and water \( (H_2O) \) as primary products. In the text, we learned about analyzing these combustion products to infer details about the original fuel mixture.
Hydrocarbon fuels are interestingly diverse, and any variation in their carbon and hydrogen content leads to different fuel types, each with unique combustion properties. Understanding the composition and combustion behavior of hydrocarbons is essential, especially in the context of environmental and engineering applications. This knowledge aids in optimizing fuel designs for better efficiency and less environmental impact.

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Most popular questions from this chapter

Ethyl acetate (A) undergoes a reaction with sodium hydroxide (B) to form sodium acetate and ethyl alcohol: The reaction is carried out at steady state in a series of stirred-tank reactors. The output from the ith reactor is the input to the \((i+1)\) st reactor. The volumetric flow rate between the reactors is constant at \(\dot{V}(\mathrm{L} / \mathrm{min}),\) and the volume of each tank is \(V(\mathrm{L})\) The concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) in the feed to the first tank are \(C_{\mathrm{A} 0}\) and \(C_{\mathrm{B} 0}(\mathrm{mol} / \mathrm{L})\). The tanks are stirred sufficiently for their contents to be uniform throughout, so that \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) in a tank equal \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) in the stream leaving that tank. The rate of reaction is given by the expression where \(k[\mathrm{L} /(\mathrm{mol} \cdot \mathrm{min})]\) is the reaction rate constant. (a) Write a material balance on \(A\) in the \(i\) th tank, and show that it yields where \(\tau=V / \dot{V}\) is the mean residence time in each tank. Then write a balance on \(\mathrm{B}\) in the ith tank and subtract the two balances, using the result to prove that (b) Use the equations derived in Part (a) to prove that and from this relation derive an equation of the form where \(\alpha, \beta,\) and \(\gamma\) are functions of \(k, C_{\mathrm{A} 0}, C_{\mathrm{B} 0}, C_{\mathrm{A}, i-1},\) and \(\tau .\) Then write the solution of this equation for \(C_{\mathrm{A} i}\) (c) Write a spreadsheet or computer program to calculate \(N\), the number of tanks needed to achieve a fractional conversion \(x_{\mathrm{A} N} \geq x_{\mathrm{Af}}\) at the outlet of the final reactor. Your program should implement the following procedure: (i) Take as input values of \(k, \dot{V}, V, C_{\mathrm{A} 0}(\mathrm{mol} / \mathrm{L}), C_{\mathrm{B} 0}(\mathrm{mol} / \mathrm{L}),\) and \(x_{\mathrm{Af}}\) (ii) Use the equation for \(C_{A i}\) derived in Part ( \(b\) ) to calculate \(C_{\mathrm{Al}}\); then calculate the corresponding fractional conversion \(x_{\mathrm{A} 1}\) (iii) Repeat the procedure to calculate \(C_{\mathrm{A} 2}\) and \(x_{\mathrm{A} 2},\) then \(C_{\mathrm{A} 3}\) and \(x_{\mathrm{A} 3},\) continuing until \(x_{\mathrm{A} i} \geq x_{\mathrm{Af}}\) Test the program supposing that the reaction is to be carried out at a temperature at which \(k=36.2 \mathrm{L} /(\mathrm{mol} \cdot \mathrm{min}),\) and that the other process variables have the following values: Use the program to calculate the required number of tanks and the final fractional conversion for the following values of the desired minimum final fractional conversion, \(x_{\mathrm{Af}}: 0.50,0.80,0.90,0.95\) 0.99, 0.999. Briefly describe the nature of the relationship between \(N\) and \(x_{\mathrm{Af}}\) and what probably happens to the process cost as the required final fractional conversion approaches \(1.0 .\) Hint: If you write a spreadsheet, it might appear in part as follows: (d) Suppose a 95\% conversion is desired. Use your program to determine how the required number of tanks varies as you increase (i) the rate constant, \(k ;\) (ii) the throughput, \(\dot{V} ;\) and (iii) the individual reactor volume, \(V\). Then briefly explain why the results you obtain make sense physically.

Methanol is produced by reacting carbon monoxide and hydrogen. A fresh feed stream containing \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) joins a recycle stream and the combined stream is fed to a reactor. The reactor outlet stream flows at a rate of \(350 \mathrm{mol} / \mathrm{min}\) and contains \(10.6 \mathrm{wt} \% \mathrm{H}_{2}, 64.0 \mathrm{wt} \% \mathrm{CO},\) and \(25.4 \mathrm{wt} \% \mathrm{CH}_{3} \mathrm{OH} .\) (Notice that those are percentages by mass, not mole percents.) This stream enters a cooler in which most of the methanol is condensed. The liquid methanol condensate is withdrawn as a product, and the gas stream leaving the condenser- -which contains \(\mathrm{CO}, \mathrm{H}_{2},\) and \(0.40 \mathrm{mole} \%\) uncondensed \(\mathrm{CH}_{3} \mathrm{OH}\) vapor \(-\mathrm{is}\) the recycle stream that combines with the fresh feed. (a) Without doing any calculations, prove that you have enough information to determine (i) the molar flow rates of CO and \(\mathrm{H}_{2}\) in the fresh feed, (ii) the production rate of liquid methanol, and (iii) the single-pass and overall conversions of carbon monoxide. Then perform the calculations. (b) After several months of operation, the flow rate of liquid methanol leaving the condenser begins to decrease. List at least three possible explanations of this behavior and state how you might check the validity of each one. (What would you measure and what would you expect to find if the explanation is valid?)

L-Serine is an amino acid that often is provided when intravenous feeding solutions are used to maintain the health of a patient. It has a molecular weight of \(105,\) is produced by fermentation and recovered and purified by crystallization at \(10^{\circ} \mathrm{C}\). Yield is enhanced by adding methanol to the system, thereby reducing serine solubility in aqueous solutions. An aqueous serine solution containing 30 wt\% serine and \(70 \%\) water is added along with methanol to a batch crystallizer that is allowed to equilibrate at \(10^{\circ} \mathrm{C}\). The resulting crystals are recovered by filtration; liquid passing through the filter is known as filtrate, and the recovered crystals may be assumed in this problem to be free of adhering filtrate. The crystals contain a mole of water for every mole of serine and are known as a monohydrate. The crystal mass recovered in a particular laboratory run is \(500 \mathrm{g},\) and the filtrate is determined to be \(2.4 \mathrm{wt} \%\) serine, \(48.8 \%\) water, and \(48.8 \%\) methanol. (a) Draw and label a flowchart for the operation and carry out a degree-of- freedom analysis. Determine the ratio of mass of methanol added per unit mass of feed. (b) The laboratory process is to be scaled to produce \(750 \mathrm{kg} / \mathrm{h}\) of product crystals. Determine the required aqueous serine solution rates of aqueous serine solution and methanol.

The respiratory process involves hemoglobin (Hgb), an iron-containing compound found in red bloodcells. In the process, carbon dioxide diffuses from tissue cells as molecular \(\mathrm{CO}_{2}\), while \(\mathrm{O}_{2}\) simultaneously enters the tissue cells. A significant fraction of the \(\mathrm{CO}_{2}\) leaving the tissue cells enters red blood cells and reacts with hemoglobin; the \(\mathrm{CO}_{2}\) that does not enter the red blood cells ( \((\mathrm{D}\) in the figure below) remains dissolved in the blood and is transported to the lungs. Some of the \(\mathrm{CO}_{2}\) entering the red blood cells reacts with hemoglobin to form a compound (Hgb. \(\mathrm{CO}_{2} ;(\) 2) in the figure). When the red blood cells reach the lungs, the Hgb.CO_ dissociates, releasing free CO_ Meanwhile, the CO_ that enters the red blood cells but does not react with hemoglobin combines with water to form carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3},\) which then dissociates into hydrogen ions and bicarbonate ions ( (3) in the figure). The bicarbonate ions diffuse out of the cells ( (4) in the figure), and the ions are transported to the lungs via the bloodstream. For adult humans, every deciliter of blood transports a total of \(1.6 \times 10^{-4}\) mol of carbon dioxide in its various forms (dissolved \(\mathrm{CO}_{2}, \mathrm{Hgb} \cdot \mathrm{CO}_{2},\) and bicarbonate ions) from tissues to the lungs under normal, resting conditions. Of the total \(\mathrm{CO}_{2}, 1.1 \times 10^{-4}\) mol are transported as bicarbonate ions. In a typical resting adult human, the heart pumps approximately 5 liters of blood per minute. You have been asked to determine how many moles of \(\mathrm{CO}_{2}\) are dissolved in blood and how many moles of \(\mathrm{Hgb} \cdot \mathrm{CO}_{2}\) are transported to the lungs during an hour's worth of breathing. (a) Draw and fully label a flowchart and do a degree-of-freedom analysis. Write the chemical reactions that occur, and generate, but do not solve, a set of independent equations relating the unknown variables on the flowchart. (b) If you have enough information to obtain a unique numerical solution, do so. If you do not have enough information, identify a specific piece/pieces of information that (if known) would allow you to solve the problem, and show that you could solve the problem if that information were known. (c) When someone loses a great deal of blood due to an injury, they "go into shock": their total blood volume is low, and carbon dioxide is not efficiently transported away from tissues. The carbon dioxide reacts with water in the tissue cells to produce very high concentrations of carbonic acid, some of which can dissociate (as shown in this problem) to produce high levels of hydrogen ions. What is the likely effect of this occurrence on the blood pH near the tissue and the tissue cells? How is this likely to affect the injured person?

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.

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