C2H5OH + O2 -> C2H4O2 +H20
This reaction drastically changes the taste of wine, if 100.0 mL of wine initially contained 12.0 g of ethanol and after a period of time 4.00 g of acetic acid were detected, what percent of the ethanol had been converted?
Isn't it just (g converted/grams initially)*100 = %?
To find the percent of ethanol that had been converted, we need to calculate the amount of ethanol remaining and compare it to the initial amount of ethanol.
The balanced chemical equation for the reaction is:
C2H5OH + O2 -> C2H4O2 + H2O
Since 1 mole of ethanol (C2H5OH) reacts to form 1 mole of acetic acid (C2H4O2), we can directly convert the given mass of acetic acid to moles:
Mass of acetic acid = 4.00 g
Molar mass of acetic acid (C2H4O2) = 60.052 g/mol
Number of moles of acetic acid = Mass / Molar mass
Number of moles of acetic acid = 4.00 g / 60.052 g/mol
Number of moles of acetic acid = 0.0666 mol
Since 1 mole of ethanol reacts to form 1 mole of acetic acid, the number of moles of ethanol that reacted is also 0.0666 mol.
Now, let's calculate the number of moles of ethanol initially present:
Mass of ethanol = 12.0 g
Molar mass of ethanol (C2H5OH) = 46.068 g/mol
Number of moles of ethanol = Mass / Molar mass
Number of moles of ethanol = 12.0 g / 46.068 g/mol
Number of moles of ethanol = 0.2606 mol
Therefore, the amount of ethanol remaining is the difference between the initial amount of ethanol and the amount reacted:
Amount of ethanol remaining = Initial amount - Amount reacted
Amount of ethanol remaining = 0.2606 mol - 0.0666 mol
Amount of ethanol remaining = 0.194 mol
To find the percent of ethanol that had been converted, we need to compare the remaining amount of ethanol to the initial amount and express it as a percentage:
Percent of ethanol converted = (Remaining amount of ethanol / Initial amount of ethanol) x 100%
Percent of ethanol converted = (0.194 mol / 0.2606 mol) x 100%
Percent of ethanol converted = 74.4%
Therefore, approximately 74.4% of the ethanol has been converted.
To find out what percentage of the ethanol had been converted, we need to determine the amount of ethanol that remained in the wine after the reaction, and then calculate the percentage decrease in the amount of ethanol.
Let's start by calculating the amount of ethanol before the reaction:
Given:
Initial volume of wine (V1) = 100.0 mL
Initial mass of ethanol (m1) = 12.0 g
Now, we need to calculate the molar mass of ethanol (C2H5OH):
Carbon (C) atomic mass = 12.01 g/mol
Hydrogen (H) atomic mass = 1.01 g/mol
Oxygen (O) atomic mass = 16.00 g/mol
Molar mass of ethanol (C2H5OH) = (2 * C) + (6 * H) + O = (2 * 12.01) + (6 * 1.01) + 16.00 ≈ 46.07 g/mol
Next, we can calculate the number of moles of ethanol before the reaction using the formula:
Number of moles (n) = mass / molar mass
Number of moles of ethanol before the reaction = m1 / molar mass = 12.0 g / 46.07 g/mol ≈ 0.260 moles
Now, to determine the amount of ethanol after the reaction, we need to understand the stoichiometry of the reaction. From the balanced equation:
C2H5OH + O2 -> C2H4O2 + H2O
We can see that for each mole of ethanol, we have an equal number of moles of acetic acid (C2H4O2) produced. Therefore, the number of moles of acetic acid (n2) formed will be equal to the number of moles of ethanol consumed.
Given:
Mass of acetic acid formed (m2) = 4.00 g
Number of moles of acetic acid formed = m2 / molar mass = 4.00 g / 60.05 g/mol ≈ 0.0666 moles (rounded to four decimal places)
Since the number of moles of acetic acid produced is equal to the number of moles of ethanol consumed, we can conclude that 0.0666 moles of ethanol were consumed in the reaction.
Therefore, the amount of ethanol remaining after the reaction is:
Number of moles of ethanol remaining = Number of moles of ethanol before the reaction - Number of moles of ethanol consumed
= 0.260 moles - 0.0666 moles ≈ 0.193 moles (rounded to three decimal places)
To determine the mass of the remaining ethanol, we multiply the number of moles by the molar mass of ethanol:
Mass of ethanol remaining = Number of moles of ethanol remaining * molar mass of ethanol
= 0.193 moles * 46.07 g/mol ≈ 8.899 g (rounded to three decimal places)
Finally, we can calculate the percentage of ethanol converted using the equation:
Percentage of ethanol converted = (Initial mass of ethanol - Mass of ethanol remaining) / Initial mass of ethanol * 100
Percentage of ethanol converted = (12.0 g - 8.899 g) / 12.0 g * 100 ≈ 26.74%
Therefore, approximately 26.74% of the ethanol was converted in this reaction.