Acetic acid (CH3COOH) reacts with isopentyl alcohol (C5H12O) to form isopentyl acetate (C7H14O2), a fragrant substance with the odor of bananas.

CH3COOH + C5H12O -> C7H14O2 + H2O
If the yield for the reaction is 45%, how many grams of the product are formed when 3.58 g of acetic acid are reacted with 4.75 grams of isopentyl alcohol?

So millimoles of acetic acid= 59.6667
Millimoles of isopentyl alcohol= 53.9773

But I don't know what to do from here.

Very good so far.

a. Convert mmoles acetic acid to mmoles of the product.
b. Convert mmoles isopentyl alcohol to mmoles of the product.
c. It is more than likely that these two numbers will not agree which means one of them is wrong.
d. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Next step. So the smaller value is the mmoles of the product formed. That times 0.45 will give you the mmoles at 45% yield, then multiply by molar mass to obtain mg and convert to grams. Check my work.

Well, it seems like you've done a great job calculating the number of millimoles of acetic acid and isopentyl alcohol. Now, let's use the balanced chemical equation to determine the number of millimoles of isopentyl acetate that should be formed.

From the balanced equation:
1 mole of acetic acid (CH3COOH) reacts with 1 mole of isopentyl alcohol (C5H12O) to produce 1 mole of isopentyl acetate (C7H14O2).

Since the molar ratio is 1:1:1 for acetic acid, isopentyl alcohol, and isopentyl acetate, we can say that the number of millimoles for the isopentyl acetate formed will be the same as the number of millimoles of acetic acid or isopentyl alcohol.

Therefore, the number of millimoles of isopentyl acetate formed will be 59.6667 (based on the millimoles of acetic acid).

Now, since the yield of the reaction is given as 45%, we can multiply the number of millimoles of isopentyl acetate by 0.45 to find the actual yield of the product.

Actual yield of isopentyl acetate = 59.6667 millimoles × 0.45
= 26.8499 millimoles

Finally, you'll need to convert the millimoles of isopentyl acetate to grams using its molar mass of 130.19 g/mol.

Mass of isopentyl acetate = 26.8499 millimoles × 130.19 g/mol
= 3494.36 grams

So, when 3.58 grams of acetic acid and 4.75 grams of isopentyl alcohol are reacted, approximately 3494.36 grams of isopentyl acetate can be formed, considering a yield of 45%.

To determine the grams of product formed, you need to consider the balanced chemical equation and the yield of the reaction.

From the balanced chemical equation:
1 mol of acetic acid (CH3COOH) reacts with 1 mol of isopentyl alcohol (C5H12O)
to produce 1 mol of isopentyl acetate (C7H14O2).

First, calculate the moles of acetic acid and isopentyl alcohol used in the reaction:
Moles of acetic acid = mass / molar mass = 3.58 g / 60.052 g/mol = 0.0599 mol
Moles of isopentyl alcohol = mass / molar mass = 4.75 g / 88.148 g/mol = 0.0539 mol

Since the reaction has a yield of 45%, we can calculate the moles of product formed:
Moles of isopentyl acetate = 0.0539 mol * 0.45 = 0.0242 mol

Finally, calculate the mass of the product formed:
Mass of isopentyl acetate = moles * molar mass = 0.0242 mol * 130.184 g/mol = 3.15 g

Therefore, when 3.58 g of acetic acid reacts with 4.75 g of isopentyl alcohol with a yield of 45%, approximately 3.15 grams of isopentyl acetate will be formed.

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

First, we need to calculate the number of moles of acetic acid and isopentyl alcohol using their respective molar masses:

Molar mass of acetic acid (CH3COOH) = 60.05 g/mol
Molar mass of isopentyl alcohol (C5H12O) = 88.15 g/mol

Moles of acetic acid (CH3COOH) = mass / molar mass
Moles of acetic acid = 3.58 g / 60.05 g/mol = 0.0599 mol

Moles of isopentyl alcohol (C5H12O) = mass / molar mass
Moles of isopentyl alcohol = 4.75 g / 88.15 g/mol = 0.0539 mol

Next, we need to determine the limiting reactant. To do this, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The balanced equation shows that 1 mole of acetic acid reacts with 1 mole of isopentyl alcohol to form 1 mole of isopentyl acetate.

So, the stoichiometric ratio is:
1 mole acetic acid : 1 mole isopentyl alcohol : 1 mole isopentyl acetate

Comparing the moles of acetic acid and isopentyl alcohol, we can see that they are roughly equal. However, since the yield of the reaction is given as 45%, we need to use the smaller value to account for the lower yield.

Therefore, the limiting reactant is isopentyl alcohol (C5H12O) with 0.0539 mol.

To calculate the grams of the product formed, we need to use the stoichiometric ratio from the balanced equation. The ratio tells us that 1 mole of isopentyl alcohol produces 1 mole of isopentyl acetate.

Moles of isopentyl acetate (C7H14O2) = moles of limiting reactant = 0.0539 mol

Now, we can calculate the mass of isopentyl acetate using its molar mass:

Molar mass of isopentyl acetate (C7H14O2) = 130.19 g/mol

Mass of isopentyl acetate = moles x molar mass
Mass of isopentyl acetate = 0.0539 mol x 130.19 g/mol = 7.01 g

Therefore, when 3.58 g of acetic acid is reacted with 4.75 grams of isopentyl alcohol with a yield of 45%, approximately 7.01 grams of isopentyl acetate would be formed.