Concept explainers
(a)
Interpretation:
Draw and label the flowchart for the process.
Concept introduction:
At steady state material balance can be written as,
Material into the system = Material out of the system.
(b)
Interpretation:
Calculate the required fresh flow rates.
Concept introduction:
At steady state material balance can be written as,
Material into the system = Material out of the system
(c)
Interpretation:
Calculate the flowrates of the air fed to the reactor, the gas leaving the reactor and the liquid water living the reactor.
Concept introduction:
At steady state material balance can be written as,
Material into the system = Material out of the system
(d)
Interpretation:
Calculate the mass flowrate of the recycle stream.
Concept introduction:
At steady state material balance can be written as,
Material into the system = Material out of the system
(e)
Interpretation:
Explain the functions of oxygen, nitrogen, catalyst, and the solvent in the process.
Concept introduction:
At steady state material balance can be written as,
Material into the system = Material out of the system
(f)
Interpretation:
Explain what can be done to improve the economics of the process.
Concept introduction:
At steady state material balance can be written as,
Material into the system = Material out of the system
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Chapter 5 Solutions
EBK ELEMENTARY PRINCIPLES OF CHEMICAL P
- The aqueous solution of acetic with a flow rate of 1000 kg/hr contains 30% acetic acid by mass and will be extracted in a countercurrent multistage process with pure isopropyl ether (IPE) to reduce the acetic acid concentration in the final raffinate phase (LN) to 2% by mass. a) Calculate the minimum solvent flow. b) Determine the theoretical number of steps required when using a solvent of 1.5 minimum solvent flow rate. c) Determine the number of theoretical steps using the McCabe-Thiele method. Chart. Acetic acid (A)-Water (B)-isopropyl ether (IPE) (C) LIQUID-Liquid Balances at 1 atm pressure and 293 K Water layer (%) Water (B) Isopropyl ether layer (%) Water (B) 0.6 Acetic acid (A) IPE (C) Acetic acid (A) IPE (C) 98.8 1.2 99.4 99.3 98.9 0.69 98.1 1.2 0.18 0.5 1.41 97.1 1.5 0.37 0.7 2.89 95.5 1.6 0.79 98.4 91.7 1.9 1.0 97.1 6.42 13.30 25.50 36.70 1.9 93.3 4.82 11.40 2.3 84.4 71.1 3.4 3.9 84.7 71.5 21.60 31.10 58.9 4.4 6.9 58.1 48.7 10.6 10.8 45.1 37.1 44.30 46.40 16.5 36.20 15.1…arrow_forwardNaturally occurring barium sulphate is converted into water–soluble formby heating with carbon, thus reducing the sulphate to barium sulphide(BaS). The resulting reaction mixture, “barium black ash”, contains 65%w/w soluble BaS and is to be leached with water. Black ash is to be fedto a tube mill at 100 tonnes per day, together with the overflow from thesecond of a cascade of thickeners, and the effluent from the mill is fed tothe first thickener. All the barium is dissolved in the mill. The strongsolution leaving the first thickener contains 20% w/w BaS by weight.Each thickener delivers a sludge with 1.5 kg liquid per kg of insolublesolid. Both liquids leaving each thickener are to have the same BaSconcentration. BaS lost with the final sludge should not exceed 1 kg perday.How many thickeners are needed?arrow_forward01:50 4G File Details 4CH003/UM1: Fundamentals of... Introduction NaBH4 OH C=0 H benzophenone diphenylmethanol Ketones are readily reduced to secondary alcohols by reaction with sodium borohydride. The aim of this experiment is to determine the time required to effect the complete conversion of the ketone, benzophenone to diphenylmethanol at room temperature. This investigation may be achieved by isolating and analyzing the reaction product after different reduction times. Infrared spectroscopy affords a convenient method of analysis; by comparing the IR spectrum of the crude reaction product at different time intervals one would expect to see the intensity of the hydroxyl absorption peak increase relative to that of the original carbonyl peak. From this you can deduce the optimum time for the reduction process at room temperature. Procedure Note the "reduction time" that the demonstrator assigns to your group. Different groups will be assigned different times, eithert = 2.5 minutes,…arrow_forward
- Suppose you have a soil that is made up of 10% organic matter with CEC = 200 cmol/kg, 40% kaolinitewith CEC = 10 cmol/kg, and 50% vermiculite with CEC = 100 cmol/kg.1- Calculate the overall CEC of the soil by taking a weighted average of the three soil components.2- What percentage of the overall CEC is contributed by the organic matter? How does that numbercompare to the 10% of the soil itself that is organic?arrow_forward17- The percentage of protein in meat products is determined by multiplying the %N as determined by the kjeldahl method by the arbitrary factor 6.25. A sample of processed meat scrap weighing 2.000 g is digested with concd H2SO4 and Hg (catalyst) until the N present has been converted to NH&HSO.. This is treated with excess NaOH, and the liberated NH3 is caught in 50.0 mL pipetful H2SO, (1.000 ml 0.01860g NazO). The excess acid requires 28.80 mL NaOH (1.000 mL = 0.1266 g KHP). Calculate % protein in the meat scrap.arrow_forward2. A solution composed of 50% ethanol (E1OH), 10% methanol (MeOH), and the rest are (H2O) is fed in weight at the rate of 100 kg/hr into a separator that produces one output stream at the rate of 60 kg/hr with the composition of 80% E1OH and 15% MeOH, and the second output stream of unknown composition. Calculate the composition (in wt. %) of the three compounds in the unknown stream and its flowrate in kg/hr. (Remember to show the variable analysis)arrow_forward
- In the common process for the manufacture of nitric acid, sodium nitrate is treated with aq. sulphuric acid containing 95% H2SO4. In order that the resulting "niter cake" may be fluid; it is desirable to use sufficient acid so that there will be 34% H2SO4 by wt in the final cake. This excess H2SO4 will actually be in combination with the Na2SO4 in the cake, forming NaHSO4, although for purposes of computation it may be assumed to be free acid. The cake will contain 1.5% water and the reaction will go to completion, but 2% of the HNO3 formed will remain in the cake. Assume that the NaNO3 used is dry and pure. Calculate per 100 kg of NaNO3 charged: (a) wt and % composition of the niter cake (b) wt of aqueous acid to be used (c) wt of water vapor and nitric acid distilled from the niter cake.arrow_forward2. In an ethanol production plant, a separator produces a 99% ethanol product from a feedstock stream containing 85% water and 15% ethanol at a rate of 450 Ib/min. The separator has two outlet streams: the ethanol product outlet stream (99% ethanol) and a residual water stream. 30% of the feedstock is bypassed and mixed with the residual water stream leaving the separator. If 60% of the ethanol entering the separator is recovered in the product stream, determine the composition of the residual water stream just after leaving the separator and the composition of the residual stream after mixing with the bypass stream.arrow_forward2. Ethanol is produced by the hydration of ethylene according to the first reaction below. However, some of the products are converted to diethyl ether according to the second reaction. C2H4 + H20 – C,H;0H 2C2H;OH - (C,H5)20 + H20 The feed to the reactor contains ethylene, steam, and N2 An effluent analysis is given in the following table: C2H4 Component %mol H2O 45.08 With a basis of 100 kmol of effluent, determine the following: CH;OH 2.95 (C2H5)20 0.27 N2 39.12 12.58 Determine the fractional yield of ethanol.arrow_forward
- One gram-mole each of CO2, 02, and N2 are fed to a batch reactor and heated to 3000 K and 5.0 atm. The two reactions given here proceed to equilibrium (also shown are the equilibrium constants at 3000 K). CO2 =CO+02 1/2 K1 = (PcoPo,)/Pco; = 0.3272 atm'/2 02 +N2 =NO K2 = PNo/(Po,PN,)"/² = 0.1222arrow_forwardB. Weight of Na2SO4 Sample 0.3512 g C. Total Vol. of Precipitant (BaCl2) used WITH 1M BaCl2 PLEASE FIND THE calculated volume of BaCl2 precipitant will be calculated using this formula: NVBAC12 = (weight of sample) (1000)(EW) V(ml)BaCl2 = (weight sample) %3D (1000)N(EW)arrow_forwardA 15.00 g sample containing mixed alkali and other inert components was dissolved and diluted to 300 mL with water. A 20 ml, aliquot was titrated with 5.02 mL of 0.5352 M HCI to reach PHP endpoint. Another 20 mL aliquot was titrated to the BCG endpoint, using up 18.87 mL of titrant in the process. Identify the alkali components and their percent weight.arrow_forward
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
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