2AL+2KOH+4H2SO4+22H2O------- 2 KAL(SO4)2+2 H2O+3H2O 1) In the given procedure fo

Question

2AL+2KOH+4H2SO4+22H2O------- 2 KAL(SO4)2+2 H2O+3H2O

1) In the given procedure for the production of alum, what is the limiting reagent?Since water is the solvent, assume that it is present in excess.

2) How much aluminum is necessary to produce 18 g of alum by way of the given procedure? This is the amount of aluminum you should use in your procedure.

3)By what factor must the mole amount of Al in the given procedure be reduced to produce 18 g of alum?

4) What is the result of scaling the number of moles of KOH used in the given procedure by the factor determined in question3? What the volume of 1.4 M KOH contains this number of moles? This is the volume of 1.4 M KOH you should use in your procedure.

5) What is the result of scaling the number of moles of H2SO4 used in the given procedure by the factor determined in question3? What volume of 9.0 M H2SO4 contains this number of moles? This is the volume of 9.0 M H2SO4 you should use in your procedure.

6) In the given procedure, a 54 mL aliquot of 10.0 M H2SO4 was used. Scale the moles of sulfuric acid in the 54 mL aliquot of 10.0 M H2So4 by the factor determined in question 3. what the volume of 9.0 M H2SO4 contains this number of moles? This the size aliquot of sulfuric acid you should use in your procedure.

7)In the given procedure, 60 mL of distilled water is used to rinse the residue obtained after gravity filtration. What is the result of scaling the 60 mL sample of distilled water by the factor determined in question 3? This is the volume of distilled water you should use in your procedure.

8) What is the result of scaling the volume of the 50/50 ethanol-water misture used in the given procedure by the factor determined in question3? This is the volume of this misture that you should use in your procedure.

Explanation / Answer

1) limiting reagent is aluminnium

2) 18/241* 26 = 1.94 gm

3)0.07 mole

4)

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