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The formation of ethyl alcohol (C2H5OH MM = 46.08 g/mol) by the fermentation of glucose (C6H12O6...

The formation of ethyl alcohol (C2H5OH MM = 46.08 g/mol) by the fermentation of glucose (C6H12O6 MM = 180.18 g/mol) may be represented by:


C6H12O6 → 2C2H5OH + 2CO2

If a particular glucose fermentation process is 87.0% efficient, how many grams of glucose would be required for the production of 56.0 g of ethyl alcohol (C2H5OH)?

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Expert Solution

Dear student,

So, 125.23 gram of C6H12O6 would be required.

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And stay safe...

With regards...

queen_honey_blossom answered 2 years ago
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