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4. The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2: C6H12O6(aq)→2C2H5OH(aq)+2CO2(g) a. How ma...

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In: Chemistry

The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2: C6H12O6(aq)→2C2H5OH(aq)+2CO2(g) a. How ma...

The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2: C6H12O6(aq)→2C2H5OH(aq)+2CO2(g)

a. How many moles of CO2 are produced when 0.310 mol of C6H12O6 reacts in this fashion?

b. How many grams of C6H12O6 are needed to form 7.90 g of C2H5OH?

c. How many grams of CO2 form when 7.90 g of C2H5OH are produced?

Solutions

Expert Solution

1 mole of C6H12O6 gives 2 moles of CO2.

1 mole of CO2 will be given by 1/2 i.e. 0.5 mole of CO2.

a. 0.310 mol of C6H12O6 will give 0.310 × 0.5 mol

= 0.155 mol

b. 1 mole of C2H5OH will be given by 0.5 mol of C6H12O6.

Number of moles = mass/molar mass

Molar mass of C2H5OH = 46.07 g/mol

Molar mass of C6H12O6 = 180.16 g/mol

Moles of C2H5OH = 7.90/46.07

= 0.171 mol C2H5OH.

Moles of C2H5OH needed = 0.171 × 2

= 0.342 mol

Mass of C6H12O6 needed = 0.342 × 180.16

= 61.62 grams

c. If 1 mole of C2H5OH is formed then 1 mole of CO2 is formed.

Moles of C2H5OH in 7.90 g = 0.171 mol

Moles of CO2 formed = 0.171 mol

Molar mass of CO2 = 44.01 g/mol

Mass of CO2 formed = 44.01 g/mol × 0.171 mol

= 7.53 grams

queen_honey_blossom answered 3 years ago Next > < Previous

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