Calculate the moles of acetic acid required to react with 36.2 grams of calcium hydroxide.

To find the moles of acetic acid necessary to react with 36.2 grams of calcium hydroxide, calculate the moles of calcium hydroxide and then apply the stoichiometric ratio from the reaction equation. The result is that 0.976 moles of acetic acid are necessary for the reaction.

To calculate the moles of acetic acid required to react with a mass of 36.2 grams of calcium hydroxide, one must first determine the molar mass of each compound. The molar mass of acetic acid (CH3COOH) is 60.05 g/mol, while the molar mass of calcium hydroxide (Ca(OH)2) is 74.09 g/mol.

Using the stoichiometry of the reaction, which is:

2 CH3COOH + Ca(OH)2
arr Ca(CH3COO)2 + 2 H2O,

we see that two moles of acetic acid react with one mole of calcium hydroxide. Therefore, the moles of acetic acid needed can be calculated by:

First, convert the mass of calcium hydroxide to moles using its molar mass:
moles of Ca(OH)2 = mass of Ca(OH)2 / molar mass of Ca(OH)2.
Then, determine the moles of acetic acid required using the stoichiometric ratio from the balanced equation:
moles of CH3COOH = moles of Ca(OH)2 × 2 (since 2 moles of acetic acid are needed per mole of calcium hydroxide).

Performing the calculations yields:

moles of Ca(OH)2 = 36.2 g / 74.09 g/mol = 0.488 moles,
moles of CH3COOH = 0.488 moles × 2 = 0.976 moles.

If you need to calculate the exact mass of acetic acid required, simply multiply the moles of acetic acid calculated (0.976 moles) by its molar mass (60.05 g/mol).