C6H12O6 (s) + NH3 (g) -> C3H8O3 (l) + C2H6O(l) +CH1.83O0.56N0.17 (s) + CO2(g) + H2O Note that 1 mole glucose produces 0.5 mol glycerol and that the ratio of NH3 to H20 is 1:1. During preparation , you accidentally add too much glucose. However, you decide to let the bioreactor run anyway. After the reaction goes to completion at least one of the starting reactants is used up, you discover that 1.4 lb-m ethanol had been produced. How much heat had been produced and subsequently removed from the system. How much glucose did you initially add to the reactor?

If the question had been stated in terms of grams, you would have had to convert grams of H₂O to moles of H₂O, then moles of H₂O to moles of O₂ (as above), and finally moles of O₂ to grams of O₂.

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2020-04-09T04:38:23+02:00

Answer:

Initially; 5.22 moles of glucose was initially added to the reactor

The heat produced and subsequently removed from the system = 6.44056 × 10 ¹¹ kJ

Explanation:

Given that the unbalanced equation of the reaction is:

[tex]C_6H_{12}O_6_{(s)}} \ + NH_3_{(g)} \ --> C_3H_8O_3_{(l)}+ \ C_2H_6O_{(l)} +CH_{1.83}O_{0.56}N_{0.17(s)}+CO_{2(g)}+ H_2O_{(l)}[/tex]

Then the balanced equation can be written as :

[tex]5.22C_6H_{12}O_6_{(s)}} \ + NH_3_{(g)} \ --> \ 2.61C_3H_8O_3_{(l)}+ \ 5.34C_2H_6O_{(l)} +5.88CH_{1.83}O_{0.56}N_{0.17(s)}+6.95CO_{2(g)}+ H_2O_{(l)}[/tex]

Thus; Initially; 5.22 moles of glucose was initially added to the reactor.

However ; the mass of ethanol = 1.4 lbm

= 635 gram (since 1 lbm = 454 gram)

From stiochiometry ;

number of moles = [tex]\frac {mass}{molar mass}[/tex]

molar mass of ethanol = 46 g/mol

so; the total numbers of moles = [tex]\frac{635 \ g}{46 \ g/mol}[/tex]

the total numbers of moles = 13.80 moles

However ; for 5.34 moles of ethanol; 5.88 moles of S. cerevisiae is formed:

Now; for 1 mole of ethanol ; we have:

[tex]\frac{5.88}{5.34}[/tex] moles of S. cerevisiae formed.

So, for 13.80 moles of ethanol, we wil have

= [tex]13.80 * \frac{5.88}{5.34}[/tex] moles of S. cerevisiae formed

= 15.19 moles of S. cerevisiae formed.

We are told that the heat of combustion of S. cerevisiae = -21.2 kJ/g

Then by the combustion of S. cerevisiae the heat produced and subsequently removed from the system = 15.19 mole × 21.2 kJ/kg ( molecular weight of S. cerevisiae )

If the molecular weight of S. cerevisiae for the largest DNA strand = 2×10⁹; Then, we have:

= 15.19 mole × 21.2 kJ/kg ( 2 × 10⁹ )

= 6.44056 × 10 ¹¹ kJ