Calculate the theoretical yield of ethyl acetate formed if 10 mL of ethyl alcohol (density = 0.79 ... (density = 0.79 G/mL; MW = 46.1 G/mol ) And 12 ML Of 17.4 M Acetic Acid Were Used .

g C2H5OH = volume x density = 10 mL x 0.79g/mL = 7.9 g.

mols C2H5OH = g/molar mass = 7.9/46.1 = approx 0.17(estimate)
g CH3COOH = volume x density = 012 mL x 17.4 M = approx 210 g (estimate)
mols CH3COOH = 209/60 = approx 3.5 (estimated)
You must go through and recalculate ALL estimates (those above AND below)
This is a limiting reagent (LR) problem.
........CH3COOH + C2H5OH ==> CH3COOC2H5 + H2O
I.........3.5.................0.17...................0........................0
So C2H5OH is the LR and 0.17 mols ethyl acetate will be formed.
Theoretical yield is 0.17 mol x molar mass ethyl acetate = ? Check all my work. Post your work if you get stuck.

Well, let's get our chemistry clown shoes on for this one!

First, we'll need to calculate the moles of ethyl alcohol and acetic acid used.

Moles of ethyl alcohol = (volume of ethyl alcohol) x (density of ethyl alcohol) / (molar mass of ethyl alcohol)
= (10 mL) x (0.79 g/mL) / (46.1 g/mol)
≈ 0.172 mol

Moles of acetic acid = (volume of acetic acid) x (molarity of acetic acid)
= (12 mL) x (17.4 mol/L)
= 0.2088 mol

Next, we'll use the balanced equation for the reaction between ethyl alcohol and acetic acid to determine the stoichiometry.

C2H5OH + CH3COOH ⟶ CH3COOC2H5 + H2O

According to the equation, 1 mole of ethyl alcohol reacts with 1 mole of acetic acid to produce 1 mole of ethyl acetate.

Since the moles of ethyl alcohol and acetic acid are equal, the theoretical yield of ethyl acetate will be 0.172 mol.

But wait, there's more! Now we need to convert the moles of ethyl acetate to grams.

The formula for ethyl acetate is C4H8O2, so its molar mass is 88.11 g/mol.

The theoretical yield of ethyl acetate = (moles of ethyl acetate) x (molar mass of ethyl acetate)
= (0.172 mol) x (88.11 g/mol)
≈ 15.15 g

So, the theoretical yield of ethyl acetate would be approximately 15.15 grams.

Remember, this calculation is just clowning around and only represents the maximum amount of ethyl acetate that could be produced. Actual experimental yields may vary due to clownish factors, such as side reactions, impurities, and experimental errors.

To calculate the theoretical yield of ethyl acetate formed, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

First, let's calculate the moles of each reactant:

Moles of ethyl alcohol (C2H5OH):
Given volume = 10 mL
Density = 0.79 g/mL
Mass = volume * density = 10 mL * 0.79 g/mL = 7.9 g
Molar mass (MW) = 46.1 g/mol
Moles = mass / MW = 7.9 g / 46.1 g/mol ≈ 0.171 mol

Moles of acetic acid (CH3COOH):
Given volume = 12 mL
Molarity (M) = 17.4 M
Moles = volume * molarity = 12 mL * 17.4 mol/L = 0.2088 mol

Now, we need to compare the moles of each reactant to determine the limiting reagent:

For the reaction: C2H5OH + CH3COOH -> CH3COOC2H5 + H2O

The stoichiometry shows that one mole of C2H5OH reacts with one mole of CH3COOH to produce one mole of CH3COOC2H5 (ethyl acetate). Therefore, the limiting reagent will be the one with the lesser number of moles.

In this case, the limiting reagent is C2H5OH, as it has fewer moles (0.171 mol) compared to CH3COOH (0.2088 mol).

The stoichiometry also tells us that the molar ratio between C2H5OH and CH3COOC2H5 is 1:1. Therefore, the moles of ethyl acetate formed will be the same as the moles of C2H5OH used.

Hence, the theoretical yield of ethyl acetate formed is approximately 0.171 mol.

To calculate the theoretical yield of ethyl acetate, we need to determine the limiting reactant in the reaction between ethyl alcohol and acetic acid. The limiting reactant is the one that limits the amount of product that can be formed.

Step 1: Calculate the moles of ethyl alcohol (C2H5OH):
Given:
- Volume of ethyl alcohol = 10 mL
- Density of ethyl alcohol = 0.79 g/mL
- Molecular weight (MW) of ethyl alcohol = 46.1 g/mol

First, convert the volume of ethyl alcohol to grams using its density:
Mass of ethyl alcohol = Volume x Density = 10 mL x 0.79 g/mL = 7.9 g

Next, convert the mass of ethyl alcohol to moles using its molecular weight:
Moles of C2H5OH = Mass / MW = 7.9 g / 46.1 g/mol = 0.171 mol

Step 2: Calculate the moles of acetic acid (CH3COOH):
Given:
- Volume of acetic acid = 12 mL
- Concentration of acetic acid = 17.4 M (moles per liter)

First, convert the volume of acetic acid to liters:
Volume of CH3COOH = 12 mL = 12 mL x (1 L / 1000 mL) = 0.012 L

Next, calculate the moles of acetic acid using its volume and concentration:
Moles of CH3COOH = Volume x Concentration = 0.012 L x 17.4 M = 0.2088 mol

Step 3: Determine the limiting reactant:
To determine the limiting reactant, compare the ratios of moles of reactants. The reactant with the smaller ratio is the limiting reactant.

The balanced chemical equation for the reaction is:

C2H5OH + CH3COOH -> CH3COOC2H5 + H2O

The stoichiometric ratio between ethyl alcohol (C2H5OH) and acetic acid (CH3COOH) is 1:1. Therefore, the ratio of moles for both reactants is the same.

Moles of C2H5OH = 0.171 mol
Moles of CH3COOH = 0.2088 mol

Since both amounts are close, we can consider 0.171 mol (ethyl alcohol) and 0.2088 mol (acetic acid) to react in a 1:1 ratio. Therefore, the limiting reactant is ethyl alcohol (C2H5OH).

Step 4: Calculate the theoretical yield of ethyl acetate (CH3COOC2H5):
Theoretical yield refers to the maximum amount of product that can be obtained from the limiting reactant.

From the balanced chemical equation, we can see that the stoichiometric ratio between ethyl alcohol (C2H5OH) and ethyl acetate (CH3COOC2H5) is also 1:1. Therefore, the moles of ethyl acetate formed will be equal to the moles of ethyl alcohol used.

Theoretical yield of CH3COOC2H5 = Moles of C2H5OH = 0.171 mol

To find the mass of ethyl acetate, multiply the moles by its molecular weight:
Molar weight of CH3COOC2H5 = MW (ethyl acetate) = 88.1 g/mol (you didn't provide this value, so I assumed it based on the molecular formula)

Mass of CH3COOC2H5 = Moles x MW = 0.171 mol x 88.1 g/mol = 15.091 g

Therefore, the theoretical yield of ethyl acetate formed is 15.091 grams.