The company is developing a \"greener\" method of making propene oxide from prop

Question

The company is developing a "greener" method of making propene oxide from propene using hydrogen peroxide, H2O2, in supercritical water with a vanadium catalyst. In the reaction 1.00 kg of vanadium catalyst and 32.0 kg of water are used for each kilogram of propene. 75 % of the vanadium catalyst can be collected and reused in a subsequent batch:

C3H6+H2O2V, H2O???>89.0% yield C3H6O+H2O

Consider the prices of the following substances.

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 58.00kg of propene oxide by this alternative green process.

Express your answer numerically in dollars using three significant figures.

Substance Price per unit Unit propene $10.97 kg hydrogen peroxide $1.96 kg vanadium catalyst $57.50 kg supercritical water $1.85 kg propene oxide $258.25 kg

Explanation / Answer

"In the reaction 1.00 kg of vanadium catalyst and 32.0 kg of water are used for each kilogram of propene." they gave the amount of catalyst & water per kg of C3H6, where as I had read in as 1 kg of C3H6O

to produce 1 kg of C3H6O, I had calculated that they needed 0.81407 kg C3H6

so, to produce 1 kg of C3H6O,
"In the reaction 1.00 kg of vanadium catalyst and 32.0 kg of water"...:for each kilogram of propene."

becomes, to produce 1 kg of C3H6O:
" In the reaction" 0.81407 "kg of vanadium catalyst" ("and 32.0 kg") (0.81407) "of water for each" 0.81407 "kg of propene."

which then to produce 1 kg of C3H6O would require::
" In the reaction" 0.81407 "kg of vanadium catalyst" 26.05kg "of water"


==========


1 mol C3H6 & 1 mol H2O2 --> 1 mol C3H6O

using molar masses, that equation becomes:
42.08 grams C3H6 & 34.015 grams H2O2 --> 58.08 grams C3H6O
which is: 42.08 kg C3H6 & 34.015 kg H2O2 --> 58.08 kg C3H6O

to produce 1 kg of C3H6O, this becomes:
42.08 / 58.08 kg C3H6 & 34.015 / 58.08 kg H2O2 --> 58.08 /58.08 kg C3H6O
which is: 0.72452 kg C3H6 & 0.5857 kg H2O2--> 1 kg C3H6O

but because the reaction gives only a 89.0 % yield,
we scale up the reactants to get that desired 1 kg of C3H6O
(0.72452 kg ) (100/89) C3H6 & (0.5857 kg) (100/89) H2O2 --> 1 kg C3H6O
which is: 0.81407 kg C3H6 & 0.65804 kg H2O2 --> 1 kg C3H6O

====================
costs per kg of C3H6O produced:
since 1.00 kg of vanadium catalyst is used for each kilogram of propene,
75 % of the vanadium catalyst can be collected,
its expense is
(25%) (0.81407kg Va cat) ($57.50 per kg) = $11.70
(0.81407 kg C3H6) ($10.97 per kg) = $8.9303
(0.65804 kg H2O2) ($1.96 per kg) = $1.2898
(26.05 kg of water) ($1.85 per kg) = $48.193
total cost of materials: $70.11 per kg C3H6O produced

================

profit: ($258.25 / kg C3H6O) - ($70.11 cost per kg) = $188.14 profit /kg

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