Answer:
( $ 8,384 )
Explanation:
- From given information we know that 1 mol mCPHA provides the oxygens to 1 mol of propene, to make 1 mole of C3H6O so:
1 mol C3H6 + 1 mol mCPHA --> 1 mol C3H6O
( Mr = 42.08 g) ( Mr = 172.57 g) ( Mr = 58.08 g )
- For 1 kg equivalent equation by dividing the whole equation by the highest Molar Mass i.e of C3H6O, the result would be:
1 mol C3H6 + 1 mol mCPHA --> 1 mol C3H6O
( 42.08 / 58.08 ) ( 172.57 / 58.08 ) ( 1 )
= ( 0.72452 kg ) ( 2.9712 kg ) ( 1 kg )
- However note that the reaction gives only a 96% yield, we scale up the reactants to get that desired 1 kg of C3H6O:
(0.72452 kg ) (96/100) + (2.9712 kg) (96/100) --> 1 kg
= ( 0.75471 kg ) + ( 3.095 kg ) ---------------> 1 kg
- The costs for each component produced:
(0.75471 kg C3H6) ($10.97 per kg) = $8.279
(3.095 kg mCPHA) ($5.28 per kg) = $16.342
(0.75471 kg C3H6) / (0.0210 kg C3H6 / L dichloromethane) = 35.939
(35.939 Litres dichloromethane) ($2.12 per L) = $ 76.19
- The cost of waste disposal ($5.00 per kilogram of propene oxide) produced total cost, disregarding labor,energy, & facility costs:
$8.279 + $16.342 + $ 76.19 + $5.00 = $105.81 per kg C3H6O produced
- Profit: ($258.25 / kg C3H6O) - ($105.81 cost per kg) = $152.44 profit /kg
- Calculate the profit from producing 55.00kg of propene oxide:
(55.00kg) ($152.44 /kg) = $8,384.2 .. ( $ 8,384 )
