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Question

Yeast converts glucose (C₆H₁₂O₆) into ethanol (d = 0.789 g/mL) in a process called fermentation. An unbalanced equation for the reaction can be written as follows:

C₆H₁₂O₆ (aq) —> C₂H₅OH (l) + CO₂ (g)

Part 1-

Balance this fermentation reaction by adding the missing coefficients.

?C₆H₁₂O₆ (aq) —> ?C₂H₅OH (l) + ?CO₂ (g)

Part 2-

If 450.0 g of glucose yields 108.4 mL of ethanol, what is the percent yield for the reaction?

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Answer 1

Hello, Kirsten,

I have a more complete breakdown, but it wouldn't fit here. I uploaded it to Google Drive: https://drive.google.com/file/d/1CkKeVpdNTTwd95TqKT_taFSCMSX8nE6z/view?usp=sharing.

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1 C6H12O6 = 2 C2H5OH + 2 CO2

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In Part 2: If 450.0 g of glucose yields 108.4 mL of ethanol, what is the percent yield for the reaction?

The molar mass is 180.2 g/mole (6*12 + 12*1 + 6*16 = 180). We have 450.0 grams (4 sig figs), so we have 2.497 moles of glucose.

We should produce 2 moles of ethanol for every mole of glucose, if everything goes well and the teacher doesn't swipe any of the ethanol.

Since we start with 2.497 moles of glucose, we should get 4.99 moles of ethanol. The molar mass of ethanol is 46.1 g/mole. If we have a perfect reaction we should produce 4.99 moles ethanol * 46.1 g/mole ethanol, or 230.2 grams.

The amount of ethanol is 108.4 ml. Multiplied by it's density (0.7892 g/mL) = 85.55 grams ethanol.

Percent yield = Actual mass/ Theoretical mass

= 85.55g/230.2g = 0.372, or 37.2 percent.

The metric term for a yield this low is "That sucks." The same term can be applied to a contributor who comes up with the wrong answer.

I hope this helps, [and is correct],

Bob

Answer 2

To balance this reaction, you need coefficient 2 in front of ethanol and coefficient of 2 in front of CO2

C6H12O6 = 2 C2H5OH + 2CO2

To Determine percentage yield, first determine theoretical yield of ethanol which is quantity you obtain by calculating from mole ratio using the grams of glucose given

1 mole of glucose/180.156gms × 450.0gm glucose = 2.50 moles of glucose

Now determine no of moles of ethanol produced theoretically using mole ratio

2.5 mole of glucose×2 moles of ethanol/1 mole of glucose = 5 moles of ethanol

convert this moles into grams

46.07 gm per/1 mol ethanol ×5 moles ethanol = 230.35 gm ethanol

Change ml of ethanol actually obtained into gms by multiplying with density .789×108.4 = 85.5276gms. This is your experimental yield.

Determine % yield

% yield = experimental yield/ theoretical yield

= 85.5276×100/230.35 = 37.13gms is your percentage yield