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 = [tex]\frac{110}{1000}\times 725\ mL[/tex] = 79.75 mL
Given that:- Density = 0.789 g/cm³ = 0.789 g/mL
So, Mass = Density*Volume = [tex]0.789\times 79.75\ g[/tex] = 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:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{62.92\ g}{46.07\ g/mol}[/tex]
[tex]Moles=1.37\ mol[/tex]
According to the reaction:-
[tex]C_6H_{12}O_6_{(aq)}\rightarrow 2C_2H_5OH_{(aq)} +2CO_2_{(g)}[/tex]
2 moles of ethyl alcohol is produced when 1 mole of glucose reacts
Also,
1.37 moles of ethyl alcohol is produced when [tex]\frac{1}{2}\times 1.37[/tex] mole of glucose reacts
Moles of glucose = 0.685 Moles
Molar mass of glucose = 180.156 g/mol
Mass = Moles*Molar mass = [tex]0.685\times 180.156\ g[/tex] = 123.41 g