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Oduvanchick [21]
3 years ago
7

A relatively simple way of estimating profit is to consider the the difference between the cost (the total spent on materials an

d waste disposal) and the earnings (the price at which the product can be sold). calculate the profit from producing 62.00 kg of propene oxide.
Chemistry
1 answer:
Maksim231197 [3]3 years ago
7 0
<span>Answer: it shows that 1mol 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 using molar masses, that equation becomes: 42.08grams C3H6 & 172.57grams mCPHA --> 58.08grams C3H6O which is: 42.08 kg C3H6 & 172.57 kg mCPHA --> 58.08 kg C3H6O to produce 1 kg of C3H6O, this becomes: 42.08 / 58.08 kg C3H6 & 172.57 / 58.08 kg mCPHA --> 58.08 /58.08 kg C3H6O which is: 0.72452 kg C3H6 & 2.9712 kg mCPHA --> 1 kg C3H6O but because the reaction gives only a 96% yield, we scale up the reactants to get that desired 1 kg of C3H6O (0.72452 kg ) (100/96) C3H6 & (2.9712 kg) (100/96) mCPHA --> 1 kg C3H6O which is: 0.75471 kg C3H6 & 3.095 kg mCPHA --> 1 kg C3H6O ========= costs per kg of C3H6O 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 Litres dichloromethane (35.939 Litres dichloromethane) ($2.12 per L) = $ 76.19 & waste disposal is $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 75.00kg of propene oxide” (75.00kg) ($152.44 /kg) = $11,433 that answer rounded off to four sig figs, is $11,430</span>
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Answer:

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The empirical formula is N₂O₅.

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.  

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.  

I like to summarize the calculations in a table.  

<u>Element</u> <u>Moles</u>  <u>Ratio¹ </u>  <u> ×2²  </u>  <u>Integers</u>³

     N        1.85    1             2             2

     O        4.63    2.503   5.005     5

¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).

²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.

³Round off the number in the ratio to integers (2 and 5).

The empirical formula is N₂O₅.

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