# chem

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?

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1. 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.

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posted by Devron
2. 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

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posted by Devron
3. COULD YOU PLS EXPLAIN IT TO ME.

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posted by shane
4. 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.

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posted by Devron

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