Hello, I need help with writing a balanced equation and seeing if my theoretical

Question

Hello, I need help with writing a balanced equation and seeing if my theoretical yield calculation is correct !

Mass data for preparation of alum

Yields of alum

CALCULATIONS:

1. Balance chemical equations for conversion of Al(s) to solid aluminum alum KAI(SO4)2 * 12H2O (s) in aqueous solution

** balance the chemical equation to obtain the correct mole to mole ration (in this case 1/1) ****

2. Theoretical Yield

1g Al /26.98g Al X 1mol Al/1mol alum X 474.39 g alum = 17.58g

is this correct?

3. Please help me provide a reason for why my experiemental yield of alum is less than my theoretical yield ???

mass of weighing paper in grams .50 g mass of weighing paper plus aluminum in grams 1.34 g mass of watch glass in grams 54.73 g mass of watch glass plus dry alum in grams 67.63 g

Explanation / Answer

1. The equations involved are :

2Al(s) + 2KOH + 6H2O2  --> K[Al(OH)4] + 3H2O

2 K[Al(OH)4] + H2S04 --> 2Al(OH)3 + 2H2O + K2SO4

2 Al(OH)3 + 3H2S04 --> Al2(SO4)3 + 6 H2O + Al2(SO4)3

Al2(SO4)3 + K2SO4 + 24 H2O --> 2K[Al(SO4)2].2H2O

Net equation :

2Al(S) + 2KOH + 4H2S04 + 22H2O --> 2K[Al(SO4)2]•12H2O + 3H2

2. Since two moles of aluminium gives two moles of alum it's mole ratio is 1:1   

Number of moles of aluminium used = (1.34 - 0.5)g / 26.98g = 0.03113 moles

As ratio is 1:1 the moles of alum produced = 0.03113 moles

Molar mass of alum = 474.39 g

Mass of alum yielded = number of moles * molar mass = 0.03113* x 474.39 = 14.769 g

Theoretical yield = 14.769 g

What you have done is wrong as for calculation you have taken 1 g of aluminium but here you should take .84 g as it is your experimental value.

3. Experimental yield is now approximately 2 g less than theoretical yield

A balanced chemical equation tells what the theoretical or ideal yield of the reaction should be, assuming perfect completion of the reaction. In reality, the actual yield may be less than the theoretical yield. Reasons for this include:

1.There may be competing reactions which hinder the process under consideration.

2.External conditions may not be perfectly maintained.

3.Reactants may not be pure.

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