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Chapter 4: Problem 95

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

If the percentage of fuel in a fuel-air mixture falls below a certain value called the lower flammability limit (LFL), which sometimes is referred to as the lower explosion limit (LEL), the mixture cannot be ignited. In addition there is an upper flammability limit (UFL), which also is known as the upper explosion limit (UEL). For example, the LFL of propane in air is 2.3 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) and the UFL is \(9.5 \%^{14}\). If the percentage of propane in a propane-air mixture is greater than \(2.3 \%\) and less than \(9.5 \%,\) the gas mixture can ignite if it is exposed to a flame or spark. A mixture of propane in air containing 4.03 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) (fuel gas) is the feed to a combustion furnace. If there is a problem in the furnace, a stream of pure air (dilution air) is added to the fuel mixture prior to the furnace inlet to make sure that ignition is not possible. (a) Draw and label a flowchart of the fuel gas-dilution air mixing unit, presuming that the gas entering the furnace contains propane at the LFL, and do the degree-of-freedom analysis. (b) If propane flows at a rate of \(150 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8} / \mathrm{s}\) in the original fuel-air mixture, what is the minimum molar flow rate of the dilution air? (c) How would the actual dilution air feed rate probably compare with the value calculated in Part (b)? (>, \(<,=\) ) Explain.

In the production of soybean oil, dried and flaked soybeans are brought into contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil. (a) Draw a flowchart of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract). (b) The soybeans contain 18.5 wt\% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to \(2.0 \mathrm{kg}\) hexane per \(\mathrm{kg}\) beans. The residual solids leaving the extraction unit contain 35.0 wt\% hexane, all of the non-oil solids that entered with the beans, and \(1.0 \%\) of the oil that entered with the beans. For a feed rate of \(1000 \mathrm{kg} / \mathrm{h}\) of dried flaked soybeans, calculate the mass flow rates of the extract and residual solids, and the composition of the extract. (c) The product soybean oil must now be separated from the extract. Sketch a flowchart with two units, the extraction unit from Parts (a) and (b) and the unit separating soybean oil from hexane. Propose a use for the recovered hexane.

Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right),\) an important solvent and intermediate in the production of many other chemicals, is produced by bubbling chlorine gas through liquid benzene in the presence of ferric chloride catalyst. In an undesired side reaction, the product is further chlorinated to dichlorobenzene, and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene. The feed to a chlorination reactor consists of essentially pure benzene and a technical grade of chlorine gas (98 wt\% \(\mathrm{Cl}_{2}\), the balance gaseous impurities with an average molecular weight of 25.0 ). The liquid output from the reactor contains \(65.0 \mathrm{wt} \% \mathrm{C}_{6} \mathrm{H}_{6}, 32.0 \% \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}, 2.5 \% \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) and \(0.5 \%\) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}_{3} .\) The gaseous output contains only \(\mathrm{HCl}\) and the impurities that entered with the chlorine. (a) You wish to determine (i) the percentage by which benzene is fed in excess, (ii) the fractional conversion of benzene, (iii) the fractional yield of monochlorobenzene, and (iv) the mass ratio of the gas feed to the liquid feed. Without doing any calculations, prove that you have enough information about the process to determine these quantities. (b) Perform the calculations. (c) Why would benzene be fed in excess and the fractional conversion kept low? (d) What might be done with the gaseous effluent? (e) It is possible to use 99.9\% pure ("reagent-grade") chlorine instead of the technical grade actually used in the process. Why is this probably not done? Under what conditions might extremely pure reactants be called for in a commercial process? (Hint: Think about possible problems associated with the impurities in technical grade chemicals.)

A Claus plant converts gaseous sulfur compounds to elemental sulfur, thereby eliminating emission of sulfur into the atmosphere. The process can be especially important in the gasification of coal, which contains significant amounts of sulfur that is converted to \(\mathrm{H}_{2}\) S during gasification. In the Claus process, the \(\mathrm{H}_{2}\) S-rich product gas recovered from an acid-gas removal system following the gasifier is split, with one-third going to a furnace where the hydrogen sulfide is burned at 1 atm with a stoichiometric amount of air to form SO \(_{2}\). $$\mathrm{H}_{2} \mathrm{S}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}$$ The hot gases leave the furnace and are cooled prior to being mixed with the remainder of the \(\mathrm{H}_{2}\) S-rich gases. The mixed gas is then fed to a catalytic reactor where hydrogen sulfide and \(\mathrm{SO}_{2}\) react to form elemental sulfur. $$2 \mathrm{H}_{2} \mathrm{S}+\mathrm{SO}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{S}$$ The coal available to the gasification process is 0.6 wt\% sulfur, and you may assume that all of the sulfur is converted to \(\mathrm{H}_{2} \mathrm{S}\), which is then fed to the Claus plant. (a) Estimate the feed rate of air to the Claus plant in \(\mathrm{kg} / \mathrm{kg}\) coal. (b) While the removal of sulfur emissions to the atmosphere is environmentally beneficial, identify an environmental concern that still must be addressed with the products from the Claus plant.

A paint mixture containing \(25.0 \%\) of a pigment and the balance binders (which help the pigment stick to the surface) and solvents (which ensure that the paint stays in liquid form) sells for 18.00 dollar/kg, and a mixture containing 12.0\% sells for 10.00 dollar /kg. (a) If a paint retailer produces a blend containing \(17.0 \%\) pigment, for how much (S/kg) should it be sold to yield a 10\% profit? (b) Paint manufacturers have begun to market "low VOC" paint as a more environmentally friendly product. What are VOCs? List some ways in which paint products can be altered to lower the VOC content.
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