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: 2Ai(s) +2KOH(aq) +4H,so,(aq) +10H20() 2KAI(SO.),'12H2o(s) +3H2(g) Alum Using the data below, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant. Theoretical Yield Trial 1 9.8899 Mass Bottle mass (g) Bottle with aluminum 10.9259 pieces (g) Final product and Number g Alum 18.6106 bottle mass (g) Percent Yield Number

Explanation / Answer

2Al(s) + 2KOH + 4H2SO4 + 10H2O 2AlK (SO4)2.12H2O + 3H2(g)

In above reaction aluminum metal will be converted to alum (2AlK (SO4)2.12H2O)

Given:

Mass of bottle = 9.8899 g

Mass of bottle + aluminum piece = 10.9259 g

Mass of bottle + final product = 18.6106g

Mass of aluminum piece = (Mass of bottle + aluminum piece)- (Mass of bottle)

                                           = 10.9259 g -9.8899 g = 1.0360 g

Mass of final product = (Mass of bottle + final product) -(Mass of bottle)

                                           =18.6106g -9.8899g = 8.7207 g (actual yield)

Now calculate theoretical yield of alum

Mass of aluminum piece = 1.0360 g

Moles Al = 1.0360 g / 27.0 g/mol = 0.03837 moles Al.

According to chemical reaction 1 mol of Al will produce 1 mol of AlK (SO4)2.12H2O

So mol of 2AlK (SO4)2.12H2O = 0.03837

Molar mass of AlK (SO4)2.12H2O = 474.390 g/mol

Mass of AlK (SO4)2.12H2O = 0.03837 x 474.390 g/mol = 18.2023 g

Theoretical yield of alum is 18.2023 g

%yield = (actual yield/theoretical yield) x 100

%yield = (8.7207 g/18.2023g) x 100

Percent yield = 47.91%

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