Respuesta :
Answer:
( $ 8,384 )
Explanation:
- Â 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 )  Â
- 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      Â
- 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) Â 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 )