Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Bonus: (10 pts) Answer the following stoichiometry-related questions: 12) Write

ID: 1041828 • Letter: B

Question

Bonus: (10 pts) Answer the following stoichiometry-related questions: 12) Write the balanced equation for the reaction of acetic acid with aluminum hydroxide to form water and aluminum acetate: 13) using the equation from problem #12, determine the mass of aluminum acetate that can be made if I do this reaction with 125 grams of acetic acid and 275 grams of aluminum hydroxide. Acetic Geido.os g /nof 1255/o.osg Ino +rom mol-2.6%5 form will or gi by- o.os3 0s of produet 141.9205 14)) What is the limiting reagent in problem #137 15) ) How much of the excess reagent will be left over in problem #13 after the reaction is complete?

Explanation / Answer

Solution:

A) The balanced chemical equation is:

Al(OH)3 + 3CH3COOH -----> Al(CH3COO)3 + 3H2O

B) mass of Acetic acid used = CH3COOH = 125 g

molar mass of acetic acid = 60 g/mol

Moles of acetic acid = mass used/molar mass = 125/60 = 2.08 mol

Mass of Aluminium hydroxide = Al(OH)3 = 275 g

molar mass of Aluminium hydroxide = 78 g/mol

Moles of Aluminium hydroxide = mass used/molar mass = 275/78 = 3.52 mol

In the balanced chemical reaction,

Moles of acetic acid used for 1 mol of Aluminium hydroxide = 3

So, Moles of acetic acid used for 3.52 mol of Aluminium hydroxide = 3*3.52 = 10.57 mol

But Moles of acetic acid used are 2.08.

So, acetic acid is limiting reagent here.

In reaction, 3 moles of acetic acid gives 1 mol of aluminum acetate.

So, 2.08 mol of acetic acid will give aluminium acetate = 2.08/3 = 0.69 mol

molar mass of aluminium acetate = 204 g/mol

Mass of aluminium acetate formed = 0.69*204 = 141.4 g

Thus 141.4 g of aluminium acetate will be formed.

C) the limiting reagent here is acetic acid.

D) Moles of acetic acid used in reaction = 2.08 mol

Moles of aluminium hydroxide required = 2.08/3 = 0.69 mol

Moles of aluminum hydroxide used = 3.52

Moles of aluminum hydroxide left unreacted = 2.08 - 0.69 = 1.39 mol

Molar mass of aluminium hydroxide = 78 g/mol

Mass of aluminium hydroxide left unreacted = 1.39*78 = 108.2 g

Mass of aluminium hydroxide left unreacted =108.2 g