Respuesta :
1) It means that the wine is 15 ABV (alcohol by volume) or the ethanol in the wine solution is 15 % v/v. It has 15 ml of pure ethanol in 100 ml of wine.
2) If in 100 mL of wine we have 15 mL of alcohol, we can use that relationship to find the volume of alcohol in 1 L.
Concentration of ethanol = 15 ml of ethanol/100 mL of wine = 15 ml of ethanol/0.10 L of wine
Volume of ethanol = 1 L of wine * 15 ml of ethanol/(0.10 L of wine)
Volume of ethanol = 150 mL
3) We know that the density is 0.8 g/cm³ or 0.8 g/mL
Density = mass/volume
mass of ethanol = density of ethanol * volume of ethanol
mass of ethanol = 0.8 g/mL * 150 mL
mass of ethanol = 120 g
4) The fermentation reaction that is used to produce etanol in wine is this one:
glucose ---> ethanol + carbon dioxide
C₆H₁₂O₆ ------> 2 C₂H₅OH + 2 CO₂
The first step to find the answer is to convert the mass of ethanol into moles of it using the molar mass.
atomic mass of C = 12.01 amu
atomic mass of H = 1.01 amu
atomic mass of O = 16.00 amu
molar mass of ethanol = 2 * 12.01 + 6 * 1.01 + 1 * 16.00
molar mass of ethanol = 46.08 g/mol
moles of ethanol = mass of ethanol /(molar mass of ethanol)
moles of ethanol = 120 g/(46.08 g/mol)
moles of ethanol = 2.60 moles
Let's consider the % yield.
% yield = actual moles of ethanol/(theoretical moles of ethanol) * 100
theoretical moles of ethanol = actual moles of ethanol/(% yield) * 100
theoretical moles of ethanol = 2.60 moles/75 * 100
theoretical moles of ethanol = 3.47 moles
C₆H₁₂O₆ ------> 2 C₂H₅OH + 2 CO₂
According to the equation of the reaction 2 moles of ethanol are produced by 1 mol of glucose. We will use that conversion to find the moles of glucose necessary to produce 3.47 moles of ethanol.
moles of glucose = 3.47 moles of C₂H₅OH * 1 mol of C₆H₁₂O₆/(2 moles of C₂H₅OH)
moles of glucose = 1.735 moles
And finally we can convert those moles into grams using the molar mass of glucose (C₆H₁₂O₆).
atomic mass of C = 12.01 amu
atomic mass of H = 1.01 amu
atomic mass of O = 16.00 amu
molar mass of glucose = 6 * 12.01 + 12 * 1.01 + 6 * 16.00
molar mass of glucose = 180.18 g/mol
mass of glucose = moles of glucose * molar mass of glucose
mass of glucose = 1.735 moles * 180.18 g/mol
mass of glucose = 313 g
Answer: 313 g of glucose should be used.
