Consider the fermentation reaction of glucose: C6H12O6 ----) 2C2H5OH+2CO2


A 1.00-mol sample of C6H12O6 was placed in a vat with excess yeast. If 35.1 g of C2H5OH was obtained, what was the percent yield of C2H5OH?

1 mole of C6H12O6=2 moles of C2H5OH

moles of C2H5OH*(46.06867g of C2H5OH/mole)=g of C2H5OH

(g of C2H5OH/35.1 g of C2H5OH)*100= percent yield of C2H5OH

Answer should have three significant figures.

COULD YOU PLS EXPLAIN IT TO ME.

Opps, typo.

Last part should say (35.1 g of C2H5OH/g of C2H5OH)*100= percent yield of C2H5OH

Answer should have three significant figures

***** Note: g of C2H5OH=46.1g

Okay, looking at the reaction that you provided,

1 mole of C6H12O6=2 moles of C2H5OH

You placed one mole of C6H12O6 in the vat. Your reaction tells you that you will get 2 moles of C2H5OH for every 1 mole of C6H12O6.

C6H12O6 ----> 2C2H5OH+2CO2

We know that 1 mole of a substance=molecular mass, so two moles of C2H5OH= 92.2g of C2H5OH produced in theory.

The precent yield=(actual yield/theoretical yield)*100

The precent yield=(35.1g/92.2g)*100=38.1%

****Sorry about the note in my last post; Forgot that it was a 1 to 2 mole ratio.

fermentation is chemical decomposition in which glucose is converted into ethyl alcohol and carbon dioxide then what will be the amount of ethyl alcohol in grams and moles, which can be obtained by fermentation of 5000 grams of glucose.

Well, looks like glucose wanted to become a party animal and turned into a bunch of C2H5OH and CO2! Let's calculate the percent yield of C2H5OH from this wild fermentation reaction.

First, we need to find the molar mass of C2H5OH (or ethanol), which is approximately 46 g/mol. Now, let's convert the mass of C2H5OH obtained (35.1 g) into moles:

moles of C2H5OH = mass of C2H5OH / molar mass of C2H5OH
moles of C2H5OH = 35.1 g / 46 g/mol

Next, we can compare this to the theoretical yield of C2H5OH. From the balanced equation, we see that for every 1 mole of glucose, we should get 2 moles of C2H5OH. Since we started with a 1.00-mol sample of C6H12O6, the theoretical yield of C2H5OH would be:

moles of C2H5OH (theoretical) = 2 moles of C2H5OH / 1 mole of C6H12O6
moles of C2H5OH (theoretical) = 2 * 1.00 mol

Finally, we can calculate the percent yield:

percent yield = (moles of C2H5OH obtained / moles of C2H5OH (theoretical)) * 100
percent yield = (35.1 g / 46 g/mol) / (2 * 1.00 mol) * 100

Now, I'll leave the final calculation to you, because math isn't one of my strong suits. Good luck!

To find the percent yield of C2H5OH, we need to compare the actual yield of C2H5OH with the theoretical yield. The theoretical yield is the maximum amount of C2H5OH that could be obtained based on the stoichiometry of the reaction.

First, we need to calculate the molar mass of C2H5OH.
C = 12.01 g/mol
H = 1.01 g/mol (x5)
O = 16.00 g/mol
Total molar mass of C2H5OH = (2 * 12.01) + (6 * 1.01) + 16.00 = 46.07 g/mol

Next, we need to convert the given mass of C2H5OH into moles.
Given mass = 35.1 g
Molar mass of C2H5OH = 46.07 g/mol
Moles of C2H5OH = Given mass / Molar mass = 35.1 g / 46.07 g/mol = 0.7612 mol

Looking at the balanced equation, we can see that 1 mole of C6H12O6 produces 2 moles of C2H5OH. Therefore, the theoretical yield of C2H5OH can be calculated as follows:

Theoretical yield of C2H5OH = 0.7612 mol * (2 moles C2H5OH / 1 mole C6H12O6) = 1.5224 mol

Now we can calculate the percent yield.

Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (35.1 g / ((1.5224 mol * 46.07 g/mol)) * 100

To find the percent yield, we need the actual yield of C2H5OH in grams. However, in the given information, only the mass of C2H5OH is provided. We need to convert moles of C2H5OH to grams using its molar mass.

Molar mass of C2H5OH = 46.07 g/mol
Actual yield of C2H5OH = Moles of C2H5OH * Molar mass of C2H5OH = 0.7612 mol * 46.07 g/mol = 35.09 g

Now we can calculate the percent yield:

Percent yield = (35.1 g / 35.09 g) * 100 = 99.97%

Therefore, the percent yield of C2H5OH is approximately 99.97%.