# Ethanol can also be manufactured from glucose, C6H12O6.

C6H12O6 2CO2 + 2C2H5OH.

A solution containing 18 kg of glucose makes only 0.92 kg of ethanol.

Calculate the percentage yield of ethanol.

## 100%

## To calculate the percentage yield of ethanol, we need to compare the actual yield of ethanol to the theoretical yield.

From the balanced chemical equation:

1 mole of glucose (C6H12O6) produces 2 moles of ethanol (C2H5OH).

First, we need to calculate the number of moles of glucose and ethanol in the given quantities:

Molar mass of glucose (C6H12O6) = 180.16 g/mol

Molar mass of ethanol (C2H5OH) = 46.07 g/mol

Number of moles of glucose:

Mass of glucose = 18 kg = 18000 g

Number of moles of glucose = Mass of glucose / Molar mass of glucose

= 18000 g / 180.16 g/mol

= 99.96 mol

Number of moles of ethanol:

Mass of ethanol = 0.92 kg = 920 g

Number of moles of ethanol = Mass of ethanol / Molar mass of ethanol

= 920 g / 46.07 g/mol

= 19.97 mol

Now, we can calculate the theoretical yield of ethanol:

Theoretical yield of ethanol = (Number of moles of glucose) * (Ratio of moles of ethanol to moles of glucose)

= 99.96 mol * (2 mol ethanol / 1 mol glucose)

= 199.92 mol

Finally, we can calculate the percentage yield of ethanol:

Percentage yield of ethanol = (Actual yield of ethanol / Theoretical yield of ethanol) * 100

= (19.97 mol / 199.92 mol) * 100

= 9.98%

Therefore, the percentage yield of ethanol is approximately 9.98%.

## To calculate the percentage yield of ethanol, we need to compare the actual yield (0.92 kg) to the theoretical yield (the maximum amount of ethanol that can be produced from 18 kg of glucose).

When 1 mole of glucose (C6H12O6) is converted to ethanol (C2H5OH), it produces 2 moles of ethanol. From the balanced chemical equation, we can see that 1 mole of glucose has a molar mass of 180 g/mol, and 1 mole of ethanol has a molar mass of 46 g/mol.

1 mole of glucose = 180 g

2 moles of ethanol = 2 x 46 g = 92 g

Now let's calculate the theoretical yield of ethanol:

Convert the mass of glucose (18 kg) to moles:

18 kg x 1000 g/kg / 180 g/mol = 100 moles

Since 1 mole of glucose produces 2 moles of ethanol, the theoretical yield of ethanol would be:

100 moles x 2 moles/1 mole = 200 moles

Convert the moles of ethanol to mass:

200 moles x 46 g/mol = 9200 g or 9.2 kg

Now we can calculate the percentage yield:

Percentage Yield = (Actual Yield/Theoretical Yield) x 100

Percentage Yield = (0.92 kg/9.2 kg) x 100

Percentage Yield = 0.10 x 100

Percentage Yield = 10%

Therefore, the percentage yield of ethanol is 10%.