Alum is a compound used in a variety of applications including cosmetics, water

Question

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be synthesized from aluminum metal, sulfuric acid, water, and potassium hydroxide, as seen in the following equation:

2Al(g) + 2KOH(aq) + 4H2SO4(aq) + 10H2O =====> 2KAl(SO4) + 12(H2O) = 3H2

Using the data below, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant.

Bottle Mass - 9.8981g

Bottle with Aluminum Pieces - 10.8955g

Final Product with Bottle Mass - 19.0414g

Explanation / Answer

The amount of Al is 10.8955 g - 9.8981 g = 0.9974 g. The atomic mass of Al is 26.98 g/mole so you have 3.697 * 10^(-2) moles.

From the equation, 2 moles of Al makes 2 moles of alum, KAL(SO4)2 so the theoretical yield is 3.697 * 10^(-2) moles.

The actual yield is mass of final product minus the mass of the bottle = 19.0414 g - 9.8981 g = 9.1433 g.

The molecular weight of alum 2KAL(SO4)2.12H2O is 474.39 g/mole (the 12H2O is there because alum typically exists as a hydrated compound. Since the reaction is shown with water as a reactant, I'm going to assume the alum is supposed to be the hydrated form).

So, the theoretical yield is the moles of alum you should have produced multiplied by its molecular weight = 3.697 * 10^(-2) moles * 474.39 g/mole = 17.54 g

% yield = 9.1433 g / 17.54 g = 52.13%

Now, if the alum you created was the unhydrated compound then the formula is KAL(SO4)2 with a molecular weight of 258.20 g/mole. If you should have created 3.697 * 10^(-2) moles from our previous calculation, then the theoretical yield is 3.697 * 10^(-2) moles * 258.20 g/mole = 9.54 g.

The % yield is then 9.1433 g / 9.54 g = 95.84 %

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