contestada

Glucose (C6H12O6)(C6H12O6) can be fermented to yield ethanol (CH3CH2OH)(CH3CH2OH) and carbon dioxide (CO2).

C6H12O6⟶2CH3CH2OH+2CO2

The molar mass of glucose is 180.15 g/mol,180.15 g/mol, the molar mass of ethanol is 46.08 g/mol,46.08 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol.

a) What is the theoretical yield (in grams) of ethanol from the fermentation of 97.5 g of glucose?

b) If the reaction produced 23.4 g of ethanol, what was the percent yield?

Respuesta :

Answer: a) 49.8 gram

b) 47.0 %

Explanation:

First we have to calculate the moles of glucose

[tex]\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{97.5g}{180.15g/mole}=0.54moles[/tex]

The balanced chemical reaction will be,

[tex]C_6H_{12}O_6\rightarrow 2CH_3CH_2OH+2CO_2[/tex]

From the balanced reaction, we conclude that

As,1 mole of glucose produce = 2 moles of ethanol

So, 0.54 moles of glucose will produce =  [tex]\frac{2}{1}\times 0.54=1.08[/tex] mole of ethanol

Now we have to calculate the mass of ethanol produced

[tex]\text{Mass of ethanol}=\text{Moles of ethanol}\times \text{Molar mass of ethanol}[/tex]

[tex]\text{Mass of ethanol}=(1.08mole)\times (46.08g/mole)=49.8g[/tex]

Now we have to calculate the percent yield of ethanol

[tex]\%\text{ yield of ethanol}=\frac{\text{Actual yield}}{\text{Theoretical yield }}\times 100=\frac{23.4g}{49.8g}\times 100=47.0\%[/tex]

Therefore, the percent yield is 47.0 %