Question

Constants I Periodic Table A fictional company has been producing propene oxide, C3H60, from propene, C3H6, with a 96.0% yield. The process requires the use of a compound called m-chloroperoxybenzoic acid, C7H5O3C1, which can be abbreviated as mCPBA. The entire reaction takes place in the solvent dichloromethane, CH, Cl2: CH,CL C3H6 + C H5O3 Cl — 96.0% yield C3H2O + C,H,O, CI The initial propene concentration in the solvent is 21.0 g/L. Consider the prices of the following substances Substance propene dichloromethane Price per unit Unit $10.97 kg $2.12 L $5.28 kg $258.25 kg mCPBA propene oxide Also consider that the cost of waste disposal is $5.00 per kilogram of propene oxide produced. Part A A relatively simple way of estimating profit is to consider the the difference between the cost (the total spent on materials and waste disposal) and the earnings (the price at which the product can be sold).Calculate the profit from producing 26.00 kg of propene oxide. Express your answer numerically in dollars using three significant figures. View Available Hint(s) 10 AED * o 2 ? Submit

Answer 1

Part A

CH,Cl2 C3H8 + C H503C1 C3H2O + C7H50,CI 96% yield

the molecular weight of propane oxide is 12.011x3 + 1.008x6 + 15.999 = 58.079

the molecular weight of propane is = 12.011x3 + 1.008x6 = 42.08

the molecular weight of mCPBA is = 12.011x7 + 1.008x5 + 15.999x3 + 35.453 = 172.566

to produce 26 kg of propene oxide the amount of propane needed = (42.08/58.079)x(100/96)x26 kg= 19.6227 kg

the cost of propane needed is 19.612x10.97 $ = 215.261$

initial propane concentration is 21g/L

so the amount of CH2Cl2 needed is 19.6227/21x1000 L = 934.414 L

cost of CH2Cl2 needed is = 934.414 Lx2.12 $ = 1980.958 $

the amount of mCPBA needed is (172.566/58.079) x (100/96) x 26 kg= 80.471 kg

cost of mCPBA needed is = 80.471 x5.28 $ = 424.885 $

total cost for waste disposal is $5x26 = $130

total cost for production = $(215.261 + 1980.958 + 424.885 + 130) = $2751.1047 for 26 kg

total earning from 26 kg of propane oxide = 258.25 x 26 = $6714.5

total profit = $6714.5 - $2751.1047 = $3963.395

Part B

C3H + H202 V, H20 - 89.0% yield C3H50 + H20

since the yield is 89% to produce 26 kg of propene oxide the amount of propane needed = (42.08/58.079)x(100/89)x26 kg= 21.166 kg

the cost of propane needed is 21.166 x10.97 = $232.191

since 25% of catalyst loaded is being used and remaining is being recovered the neat amount of catalyst required is = 21.166x25/100 = 5.2915 kg

the cost of vanadium catalyst needed is 5.2915 x57.50 = $304.261

the molecular weight of hydrogen peroxide is 34.0147 g/mol

the amount of hydrogen peroxide needed is (34.0147/58.079) x (100/89) x 26 kg = 17.109 kg

the cost of hydrogen peroxide needed is 17.109 x 1.96 = $33.534

the amount of supercritical water needed is = 21.166x32 = 677.312 kg (32 kg water for 1 kg propane)

the cost of supercritical water needed is = 677.312x1.85 = $1253.027

total costing for 26 kg of propane oxide = 232.191+304.261+33.534+1253.027 = $1823.0132

total profit for 26 kg = 6714.5-1823.0132 = $4891.487