acetic acid and isopentyl alcohol form isopentyl acetate and water

The excess reagent is acetic acid, and the limiting reagent is isopentyl alcohol.

How great is the molar excess (how many times greater)?

To determine the molar excess, you need to compare the molar ratio of the excess reagent (acetic acid) to the limiting reagent (isopentyl alcohol) in the balanced chemical equation.

The balanced equation for the reaction is:

acetic acid + isopentyl alcohol → isopentyl acetate + water

The stoichiometric coefficients in this equation represent the molar ratios. From the equation, you can see that for every 1 mole of isopentyl alcohol, 1 mole of acetic acid is required.

Now, let's assume you have excess acetic acid and a certain amount of isopentyl alcohol. To find the molar excess, you need to compare the number of moles of acetic acid present to the number of moles of isopentyl alcohol used.

1. Start by determining the number of moles of isopentyl alcohol used in the reaction. You can do this by using the given mass or volume of isopentyl alcohol and its molar mass.

2. Next, calculate the number of moles of acetic acid present. You can use the given mass or volume of acetic acid and its molar mass.

3. Finally, divide the number of moles of acetic acid by the number of moles of isopentyl alcohol to find the molar excess. This ratio will tell you how many times greater the acetic acid is compared to the isopentyl alcohol.

As an AI, I don't have access to specific values or information from the question. You'll need to provide the specific amounts of acetic acid and isopentyl alcohol, such as their masses or volumes, in order to calculate the molar excess precisely.