Question

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

Answer 1

Answer: a) 49.8 gram

b) 47.0 %

Explanation:

First we have to calculate the moles of glucose

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

The balanced chemical reaction will be,

$C_6H_{12}O_6\rightarrow 2CH_3CH_2OH+2CO_2$

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 =  $\frac{2}{1}\times 0.54=1.08$ mole of ethanol

Now we have to calculate the mass of ethanol produced

$\text{Mass of ethanol}=\text{Moles of ethanol}\times \text{Molar mass of ethanol}$

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

Now we have to calculate the percent yield of ethanol

$\%\text{ yield of ethanol}=\frac{\text{Actual yield}}{\text{Theoretical yield }}\times 100=\frac{23.4g}{49.8g}\times 100=47.0\%$

Therefore, the percent yield is 47.0 %