Synthesis of Alum

Experiment 2 Online Tutorial >> Synthesis of Alum >> Tutor 2

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Experiment 2 - Synthesis of Alum

Tutor 2 - Calculate the volume of solution required to dissolve a sample

In this tutor we will calculate the volume of KOH solution required to dissolve a 1 gram sample of aluminum.

The sample of aluminum from the can weighs about 1 g. Assume that the entire sample is aluminum and calculate the volume of 1.5M KOH solution (in milliliters) required to completely dissolve the sample. (Please give your answer to 2 significant figures)

Hint

mL 1.5 M KOH

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Hint:

If you are having trouble, scroll below to get step by step help in solving this problem.

The following tutors are a step by step walkthrough that show the details on how to solve the above problem.

In the previous tutor we determined the balanced reaction for the dissolution of Aluminium in KOH solution.

Hint

2Al(s) + 2KOH(aq) + 6H2O(l) → 2Al(OH)4-(aq) + 2K+(aq) + 3H2(g)

Based on the balanced reaction above, what is the relationship between the moles of KOH consumed and moles of Al consumed?

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mole(s) of KOH consumed for each 1 mole of Al

Correct. There is 1 mole of KOH consumed for each mole of Al that is consumed.

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Hint:

Based upon the balanced equation, there are 2 moles of KOH consumed for each 2 moles of Al consumed.

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Hint:

This ratio can be reduced to 1:1. There is 1 mole of KOH consumed for each mole of Al consumed.

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Hint:

To balance the number of sulfate ions in the reaction, start by entering 1.

Hint:

Because there is 1 atom of Pb on the left and 1 sulfate ion on the left, enter 1 to balance the products with reactants in the reaction.

Hint:

Based upon the balanced equation, 1 mole sulfate is consumed for each 1 mole of Pb2+ consumed.

Our next task is to determine the number of moles present in our sample of aluminum. How many moles are there in 1 g of Al? (Note: the atomic mass of Al is 26.982 g/mol)

Hint

moles Al

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How do you convert from grams of Al to moles of Al?

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Hint:

To convert from grams to moles, use the atomic mass of Al (26.982 g/mol).

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Hint:

Assuming the sample is 100% Al, the result is:
(1g Al) / 26.982 g/mol Al) = 0.03706 moles Al.

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How many moles of potassium hydroxide are required to react with the calculated number of moles of Al in our 1 g sample?

Hint

moles KOH

Our last step is to convert the moles of potassium hydroxide calculated above to volume required. How many mL of a 1.5M KOH solution are required to react with the 1 g Al sample?

mL of 1.5M KOH

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Hint:

From the balanced chemical equation, we see that 1 mole of KOH is required to react with each 1 mole of Al. Using this 1:1 ratio, can you determine how many moles of KOH are required?

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Hint:

0.03706 moles Al * (1 mole KOH consumed) / (1 mole Al consumed)
= 0.03706 moles KOH.

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Hint:

To convert from moles of KOH to mL of KOH, use the molarity of potassium hydroxide, KOH (1.5M). What are the units of molarity (M)?

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Hint:

Molarity = moles/volume in L so volume (in L) = mol/M.
Using this conversion, how many mL are required?

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Hint:

The result is: 0.03706 moles / 1.5 M KOH
= 0.02471L = 24.71 mL or 25 mL if using about 1 g of Al.

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