Question

Most wine is prepared by the fermentation of the glucose in grape juice by yeast: C6H12O6(aq) --> 2C2H5OH(aq) + 2CO2(g) How many grams of glucose should there be in grape juice to produce 725 mLs of wine that is 11.0% ethyl alcohol, C2H5OH (d=0.789 g/cm3), by volume?

Answer 1

Answer:

123.41 g

Explanation:

Given that the ethyl alcohol produced is 11.0 % by volume.

It means that 1000 mL contains 110 mL of ethyl alcohol

Given that the volume is:- 725 mL

So,

Volume of ethyl alcohol = $\frac{110}{1000}\times 725\ mL$ = 79.75 mL

Given that:- Density = 0.789 g/cm³ = 0.789 g/mL

So, Mass = Density*Volume = $0.789\times 79.75\ g$ = 62.92 g

Calculation of the moles of ethyl alcohol as:-

Molar mass of ethyl alcohol = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{62.92\ g}{46.07\ g/mol}$

$Moles=1.37\ mol$

According to the reaction:-

$C_6H_{12}O_6_{(aq)}\rightarrow 2C_2H_5OH_{(aq)} +2CO_2_{(g)}$

2 moles of ethyl alcohol is produced when 1 mole of glucose reacts

Also,

1.37 moles of ethyl alcohol is produced when $\frac{1}{2}\times 1.37$ mole of glucose reacts

Moles of glucose = 0.685 Moles

Molar mass of glucose = 180.156 g/mol

Mass = Moles*Molar mass = $0.685\times 180.156\ g$ = 123.41 g