Answer to Question #254530 in Chemistry for Katerina

Question

a) Write the balanced equation for the reaction of acetic acid with aluminium hydroxide to form water and aluminium acetate:

b) Using the equation determine the mass of aluminium acetate that can be made if we start the reaction with 150 grams of acetic acid and 300 grams of aluminium hydroxide.

c) What is the limiting reagent?

d) How much of the excess reagent will be left over after the reaction is complete?

Answer 1

acetic acid: CH₃COOH

aluminium hydroxide: Al(OH)₃

water: H₂O

aluminium acetate: Al(CH₃COO)₃

Solution:

(a):

Balanced chemical equation:

3CH₃COOH + Al(OH)₃ → 3H₂O + Al(CH₃COO)₃

(b-c):

The molar mass of CH₃COOH is 60 g/mol

The molar mass of Al(OH)₃ is 78 g/mol

Determine moles of each reactant:

(150 g CH₃COOH) × (1 mol CH₃COOH / 60 g CH₃COOH) = 2.500 mol CH₃COOH

(300 g Al(OH)₃) × (1 mol Al(OH)₃ / 78 g Al(OH)₃) = 3.846 mol Al(OH)₃

According to the chemical equation above:

3 mol of CH₃COOH reacts with 1 mol of Al(OH)₃

Thus, 2.500 mol of CH₃COOH reacts with:

(2.500 mol CH₃COOH) × (1 mol Al(OH)₃ / 3 mol CH₃COOH) = 0.833 mol Al(OH)₃

However, initially there is 3.846 mol of Al(OH)₃ (according to the task).

Therefore, CH₃COOH acts as limiting reagent and Al(OH)₃ is excess reagent.

According to stoichiometry:

3 mol of CH₃COOH produces 1 mol of Al(CH₃COO)₃

Thus, 2.500 mol of CH₃COOH produces:

(2.500 mol CH₃COOH) × (1 mol Al(CH₃COO)₃ / 3 mol CH₃COOH) = 0.833 mol Al(CH₃COO)₃

The molar mass of Al(CH₃COO)₃ is 204 g/mol

Therefore,

(0.833 mol aluminium acetate) × (204 g aluminium acetate / 1 mol aluminium acetate) = 170 g

170 grams of Al(CH₃COO)₃ can be made.

(d):

Al(OH)₃ is excess reagent

(3.846 mol Al(OH)₃ - 0.833 mol Al(OH)₃) = 3.013 mol Al(OH)₃

The molar mass of Al(OH)₃ is 78 g/mol

Therefore,

(3.013 mol Al(OH)₃) × (78 g Al(OH)₃ / 1 mol Al(OH)₃) = 235 g Al(OH)₃

235 grams of Al(OH)₃ will be left over after the reaction is complete.

Answer:

(a): 3CH₃COOH + Al(OH)₃ → 3H₂O + Al(CH₃COO)₃

(b): 170 g of Al(CH₃COO)₃

(c): CH₃COOH is the limiting reagent

(d): 235 g of Al(OH)₃