Name Lab Section PREPARATION OF POTASSIUM ALUMINUM SULFATE DoDECAHYDRATE Date 0.

Question

Name Lab Section PREPARATION OF POTASSIUM ALUMINUM SULFATE DoDECAHYDRATE Date 0.g98 1. Mass of aluminum used 2. Mass of ALUM obtained 29514 Calculations Calculate the theoretical yield, grams of ALUM possible, from the mass of aluminum used in 3. this preparation. Show a complete calculation string, including units, with the following space. final value reported to the proper number of significant figures, in the 0.298a 411 1 mo 26.98Kg Ind AI | Inn( alum Calculate the % yield for this reaction using the formula given. Show a complete calculation string, including units, with the final value reported to the proper number of significant figures, in the following space. 4, space.wiedzGAs1gxin-82. 0%td 12.15 19 x 100 geld = 1.2.457g 82.0 %Yield = Leoretical x 100 1.la d x100 % und %Yield /s Lab Related Calculations: Show a complete calculation string, including units, with the final value reported to the prope r number of significant figures, on the back of this page. 1. What mass of hydrogen gas may be produced by the reaction of 1.00 grams of aluminum with excess potassium hydroxide? This was the first step in the procedure. 2. What quantity of sulfuric acid, in moles, is required to produce 4.74 grams of alum? 3, How many grams of alum would be produced can? Assume a can weight of 18.7 grams. Rev D- 8-2017 by the reaction of a complete beverage Page 5 of 5

Explanation / Answer

3. Chemical equation for Al to alum formation,

2Al(s) + 2KOH + 4H2SO4 + 22H2O ---> 2K[Al(SO4)2].12H2O + 3H2

So, 1 mol of Al produces 1 mol of alum

moles of Al taken = 0.898 g/27 g/mol = 0.03326 mol

molar mass of alum = 474.39 g/mol

mols of alum formed = 0.03326 mol

Theoretical yield of alum = 0.03326 mol x 474.39 g/mol = 15.78 g

4. Experimental yield of alum = 12.957 g

Percent yield of alum = (Experimental yield/theoretical yield) x 100

                                   = (12.957 g/15.78 g) x 100

                                   = 82.11%

Lab related calculations

1. reaction of Al with KOH

2Al + 2KOH + 6H2O ---> 2K[Al(OH)4] + 3H2

So, 2 mol of Al gives 3 mol of H2 gas

mols of Al = 1.0 g/27 g/mol = 0.037 mol

mols of H2 formed = 0.037 mol x 3/2 = 0.0555 mol

mass of H2 gas produced = 0.0555 mol x 2.016 g/mol = 0.112 g

2. From the Al to alum formation equation given above,

1 mol of alum is formed from 2 mols of H2SO4

mols of alum = 4.74g/474.39 g/mol = 0.01 mol

mols of H2SO4 required = 2 x 0.01 mol = 0.02 mol

quantity of H2SO4 required = 0.02 mol x 98.08 g/mol = 1.962 g

3. considering beverage can is made of pure Alumnium

mols of Al = 18.7 g/27 g/mol = 0.6926 mol

mols of Al formed = 0.6926 mol

mass of alum produced = 0.6926 mol x 474.39 g/mol = 328.56 g

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