Question

Part A and B

Alternative method The company is developing a greener" method of making propene oxide from propene using hydrogen peroxide, H2O, in supercritical water with a vanadium catalyst. In the reaction 1.00 kg of vanadium catalyst and V. 1,0 320 kg of water are used for each kilogram of propene. 75% of the vanadium catalyst can be collected and reused in a subsequent batch:CH + H2O, C,H, O H , OConsider the prices of the 89.05 yield following substances Substance Price per unit Unit propene $10.97 hydrogen peroxide $1.96 vanadium catalyst $57.50 supercritical water $1.85 kg propene oxide $258.25 kg There is virtually no waste produced with this green process and there is no waste disposal fee, but the yield is only 89.0%. Part B Calculate the profit from producing 29.00 kg of propene oxide by this alternative green process. Express your answer numerically in dollars using three significant figures. View Available Hint(s) AED O ? Submit A fictional company has been producing propene oxide, C,H,O, from propene, CsHs, with a 96.0% yield. The process requires the use of a compound called m-chloroperoxybenzoic acid, C,H,O, Cl, which can be abbreviated as mCPBA. The entire reaction takes place in the solvent dichloromethane, CH, Cle: CH.CH C3H6 +C HOCl + C3H6O+C HOẠCH 96.0% yield The initial propene concentration in the solvent is 21.0 g/L. Consider the prices of the following substances. Substance propene dichloromethane mCPBA propene oxide Price per unit Unit $10.97 $2.12 $5.28 $258.25 kg 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 29.00 kg of propene oxide. Express your answer numerically in dollars using three significant figures. View Available Hint(s) % A¢* ? Submit

Answer 1

Question 1

As the weight of the final product is given, back calculations need to be made to arrive at the weight of the starting material.

Molar mass of propene oxide = 58.08 g/ mol

Mass of propene oxide = 29.00 kg

Moles of propene oxide = mass/ molar mass = (29.00 x 1000) g / 58.08 g/mol = 499.311 mol

If the yield of the reaction were 100 % then to get 499.311 mol propene oxide, 499.311 mol of propene would be required.

But the yield is 89%. So the amount of propene required is = moles of prod x 100 / yield = 499.311 mol x 100 / 89.0 = 561.024 mol

Molar mass of propene = 42.081 g/ mol

Mass of propene required = moles x molar mass = 561.024 mol x 42.081 g / mol = 23608.45 g = 23.608 kg

Moles of H2O2 = moles of propene = 561.024 mol

(Based on balanced chemical equation)

Molar mass of H2O2 = 34.05 g/mol

Mass of H2O2 = moles x molar mass = 561.024 mol x 34.05 g/mol = 19102.8g = 19.103 kg

Vanadium required 1kg for 1 kg propene. So total vanadium required = 23.608 kg

Vanadium recovered = 75%

So Vanadium consumed = 25%

Mass of vanadium consumed = 25 x 23.608 kg/ 100 = 5.902 kg

H2 O required is 32 Kg per 1 kg of propene

Total H2O required = 32 x 23.608 kg = 755.456 kg

Now the cost of materials used is equal to the quantity of material used in kg x per kg cost of material

Cost of propene = 23.608 kg x $10.97/kg = $258.980

Cost of H2O = 755.456 kg x $1.85/kg = $1397.594

Cost of H2O2 = 19.103 kg x $1.96/kg = $37.442

Cost of vanadium = 5.902 kg x $57.50/kg = $339.365

Total cost for the reaction is sum of all the starting material cost

On addition of above numbers we get cost of $2033.381

Earning on propene oxide = 29 kg x $ 258.25/kg = $7489.25

Profit = $7489.25 - $ 2033.381 = $5455.869

Question 2

Molar mass of propene oxide = 58.08 g/mol

Mass of propene oxide = 29.0 kg

Moles of propene oxide = 29000 g / 58.08 g/mol = 499.311 mol

Yield 96%

So the moles of propene = 499.311 mol x 100 / 96 = 520.116 mol

Mass of propene required = 520.116 mol x 42.081g/mol = 21886.9 g = 21.887kg

Moles of mcpba = 520.116mol

Molar mass of mcpba = 172.56 g/mol

Mass of mcpba = 520.116 mol x 172.56 g/mol = 89751.2 g = 89.751 kg

1 L DCM is required for 21g of propene

For 21887 g propene, DCM required = 1 x 21887/21 = 1042.24 L

Calculating the cost for each material

Cost for propene = 21.887 kg x $10.97 = $240.100

Cost of mcpba = 89.751 kg x $5.28 = $473.885

Cost of DCM = 1042.24 x $ 2.12 = $2209.545

Total cost = $2923.530

Coat of waste = $5.00 x 29 = $145

Final expenses = $2923.530 + $145 = $3068.53

Total earnings = 29 kg x $258.25 = $7489.25

Profit = $ 7489.25 - $3068.530 = $4420.720