Synthesis of Alum

Experiment 2 Online Tutorial >> Synthesis of Alum >> Tutor 3

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Experiment 2 - Synthesis of Alum

Tutor 3 - Calculate the percentage yield of alum

The following tutor will guide you through the processes of calculating the percentage yield of alum.

In one experiment, 0.6343 g aluminum metal reacted with 50.3 mL of 1.43 M KOH to produce 4.3654 g of alum. What is the percentage yield of alum? (Please give your answer to 2 significant figures)

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To calculate percentage yield, you divide the actual yield by the theoretical yield and then multiply by 100%. To determine the theoretical yield, first you need a balanced chemical equation for the reaction and then you need to determine which of the two reactants is the limiting reagent. The calculation for theoretical yield is based on the limiting reagent.

Hint

The balanced chemical equation for the reaction is:

Al3+(aq) + K+(aq) + 2SO42-(aq) + 12H2O(l) → KAl(SO4)2 • 12H2O(s)

How many moles of each reactant are you starting with:

moles Al

moles KOH

Which reactant is the limiting reagent:

Aluminum

KOH

Correct. Al is limiting and KOH is in excess.

Hint:

We can now use the calculated values and the balanced reaction to determine the limiting reagent. In the balanced reaction, what is the stoichiometric relationship between moles of KOH and moles of Al consumed?

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Hint:

There is a 1 to 1 mole relationship, 1 mole of aluminum reacts with 1 mole of KOH. Based on this relationship, which reactant is the limiting reagent?

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Hint:

Al is the limiting reagent. 0.02351 moles Al will react with 0.02351 moles of KOH. There are, however, 0.07193 mol of KOH which is more than required. Thus Al is limiting and KOH is in excess.

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Hint:

To determine how many moles of Al were initially present, we will need to use the atomic mass of Al (26.9815 g/mol). Using this information, how many moles are there in 0.6343 g Al?

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Hint:

Moles of Al = (0.6343 g Al) / (26.9815 g/mol Al) = 0.02351 moles Al

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Hint:

To find moles of KOH recall that molarity = moles/volume, thus moles=(molarity)*(volume). Using this relation, calculate the number of moles in 50.3 mL of 1.43 M KOH.

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Hint:

Moles of KOH = (1.43 M) * (0.0503 L) = 0.0719 mol KOH

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Hint:

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Hint:

There is a 1 to 1 mole relationship, 1 mole of aluminum reacts with 1 mole of KOH. Based on this relationship, which reactant is the limiting reagent?

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Hint:

Al is the limiting reagent. 0.02351 moles Al will react with 0.02351 moles of KOH. There are, however, 0.07193 mol of KOH which is more than required. Thus Al is limiting and KOH is in excess.

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Now that we have determined that Al is the limiting reagent, how many grams of alum (potassium aluminum sulfate dodecahydrate, KAl(SO4)2 • 12H2O will be produced?

(Note: the molecular mass of KAl(SO4)2 • 12H2O is 474.39 g/mol)

Hint

grams KAl(SO4)2 • 12H2O

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From the balanced chemical equation, 1 mole of Al produces 1 mole of KAl(SO4)2 • 12H2O. Thus there is a 1 to 1 mole relationship between Al consumed and KAl(SO4)2 • 12H2O produced. How many moles of KAl(SO4)2 • 12H2O will be produced?

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Hint:

0.02351 mol of Al were consumed so 0.02351 mol of
KAl(SO₄)₂ • 12H₂O will be produced. How do you convert this value to grams?

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Hint:

Using the molecular mass of KAl(SO4)2 • 12H2O, (474.39 g/mol) the result is : (0.02351 mol Alum) * (474.39 g/mol Alum) = 11.15 g KAl(SO4)2 • 12H2O.

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What is the percentage yield of alum?

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Percentage yield = [actual yield / theoretical yield] x 100%. Using this equation can you determine percentage yield?

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Hint:

Substituting the given yield and the calculated theoretical yield into the equation gives: Percentage yield = (4.3654 g / 11.15 g) x 100% = 39.15%

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