Problem! (1 Viewer)

[Chocolat]

Hey guys...

I'm having immense trouble on a particular Chemistry problem and I would be so appreciative if anyone could help me .

Question:
Assuming 100% conversion, calculate the mass of glucose that can be produced from the complete photosynthesis of 1000 kilograms of carbon dioxide.

I have gotten either 148 kg or 696kg because I'm unsure about a particular step. Please let me know what you got and if it's something different, how.

THANK YOU!

 

[toadstooltown]

6CO2+6H2O -> C6H12O6 + 6O2.
CO2 is limiting reagent.
Moles of CO2 = 1000000g÷(12+2×16) =22727.27moles
No of moles of CO2 is 6× that of glucos so moles of glucoes = 22727.27÷6
= 3787.87
Mass = moles × molar mass
= 3787.87 × (6×12+12 +16×6)
= 681818g
= 681.81kg

Let me know if I've made mistakes. I haven't checked it.

 

[twilight1412]

the working out seems ok to me ...
btw can you clarify something for me.. in chemistry its the working out thats important ..... so if you write you answer as 1 will it be considered a calculation error? and if you use that one to work out your next answer would that be a carry on error? would you get full marks for that?

 

[Chocolat]

Thanks so much!

And just to confirm, I was also given the questions:

Calculate the mass of ethanol that can be fermented from this amount of glucose. {I got 349kg...)

and

Calculate the mass of carbon dioxide released from the complete combustion of this amount of ethanol. {I got 666.3kg)

I would appreciate if you guys could quickly do them and just type up a response if you get a different answer and how...thanks so much!

 

[jest3r]

the working out seems ok to me ...
btw can you clarify something for me.. in chemistry its the working out thats important ..... so if you write you answer as 1 will it be considered a calculation error? and if you use that one to work out your next answer would that be a carry on error? would you get full marks for that?

I recall in my chemistry preliminary exam, I still got full marks for a particular question, even though the answer was wrong, my calculations were correct. So I guess examiners are more concerned about how you got your answer than what your answer is (not saying that you shouldn't try to get the right answer). This also applies with mathematics and physics.

 

[jest3r]

Oh, and in answer to your questions...


Calculate the mass of ethanol that can be fermented from this amount of glucose. {[You] got 349kg...)

With the overall reaction of fermentation being... (hope you can figure out where the subscripts are)

C6H1206 (aq) -> 3C2H6O (aq) + CO2 (g)

(note that fermentation requires yeast and a temperature of 37 degrees C.)

moles of glucose = mass / molecular weight

where the mass of glucose is given, as you said before, it was 349 kg, which is 349 000 g.

therefore,

moles of glucose = 349 000 / ([6 x 12] + [12 x 1] + [6 x 16])

= 1938.88 moles

moles of ethanol = 3 x 1938.88

= 5816.66 moles

mass of ethanol = moles x molecular weight

= 5816.66 x ([2 x 12] + [6 x 1] + [1 + 16])
= 267566.66 kg
= 267.56 g of ethanol

Now, for your second question...

Calculate the mass of carbon dioxide released from the complete combustion of this amount of ethanol. {[You] got 666.3kg)

Complete combustion of ethanol should be as follows...

C2H5OH (aq) + 3O2 (g) -> 2CO2 (g) + 3H2O (g)

moles of ethanol = mass / molecular weight
= 666 300 / (2 x 12) + (5 x 1) + (1 + 16)
= 14806 moles

moles of CO2 = 2 x 14806.66
= 29613.33 moles

moles of gas = v / 22.4

volume of gas = moles x 22.4

volume of carbon dioxide = 29613.33 x 22.4
= 663331.2 mL or g (assuming that, experimentally, this is being carried out at normal sea level)

= 663.33 kg

Hopefuly this answers your questions correctly...give me a shout if I've done something, or the whole thing wrong. It feels like ages ago since I last did stoichiometry problems.

 

[yoakim]

6CO2+6H2O -> C6H12O6 + 6O2.
CO2 is limiting reagent.
Moles of CO2 = 1000000g÷(12+2×16) =22727.27moles
No of moles of CO2 is 6× that of glucos so moles of glucoes = 22727.27÷6
= 3787.87
Mass = moles × molar mass
= 3787.87 × (6×12+12 +16×6)
= 681818g
= 681.81kg

Let me know if I've made mistakes. I haven't checked it.

To find the n.o of moles of C02, is it (12+2×16) or would it be (12x16+2×12)? <-- because C02 has 6 moles at the start which needs to be divided by 6 to be in the same ratio as C6H12O6. Is this a common mistake that I need to be aware of?

 

[tristambrown]

I have a different eqn for the fermentation

roland smith
conquering chemistry
fourth edition
p33
...............fermentation
C6H12O6 ----------> 2CO2 + 2CH3-CH2-OH
(above states 3CH3-CH2-OH)

With this eqn i get 681.82 KG Glucose (aq) (rounded to 2 dec place)

->348.48 (2 dec place) KG Ethanol (l) + 333.33 (2 dec place)KG CO2 (g)