Data: Mass of aluminum foil ____ 0.52 grams Limiting Reagent (based on observati

Question

Data: Mass of aluminum foil ____ 0.52 grams Limiting Reagent (based on observations of first part of lab) ____ A1 Mass of plastic cup ___ 2.78 grams Mass of alum crystal + plastic cup ____ 10.39 grams Data Analysis: Overall rxn: 2Al + 2KOH + 6H_2SO_4 + 22H_2O rightarrow 2(KAl(SO_4)_2 * (12H_2O) (alum) + 3H_2 Calculate the following: A Mass of Alum formed B Molar mass of alum C Support your observed limiting reagent with calculations. (Assume you started with 0.075 moles of KOH, excess H_2SO_4 and excess water. Don't forget about your starting amount of Aluminum foil). D What is your theoretical yield (in grams) of Alum? E How many grams of excess reactant remained at the end of the reaction? F Percent yield of the reaction

Explanation / Answer

(C)

Moles Al = 0.52 / 27 = 0.019 mol

From the balanced equation,

2 mol of Al requires 2 mol of KOH

Then, 0.019 mol of Al requires 0.019 mol of KOH.

Hence Al is limiting reagent.

(D)

And from the above balanced equation,

2 mol Al forms 2 mol Alum

then, 0.019 mol Al forms 0.019 mol Alum.

So,

Mass of Alum = Theoretical yield = Moles * molar mass = 0.019 * 471 = 8.95 g.

(E)

AMount of excess reagent remained = Moles * molar mass = (0.075 - 0.019) * (56) = 3.14 g.

(F)

Actual yiled of alum = 10.89 - 2.78 = 8.11 g.

Percent yield = (actual yield * 100) / theoretical yield = (8.11 * 100) / 8.95 = 90.6 %

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