Question

-Please Check my calculations-

Mass Data for preparation of Alum

Mass Weighing paper: 0.3832 grams

Mass Weighing paper plus Aluminum: 1.3838 grams

Mass of watch glass: 56.5001 grams

Mass of watch glass plus dry alum: 68.8550

Balanced Equation: 2Al(s)+2KOH(aq)+4H2SO4(aq)+22H2O(l)=2KAl(SO4)2*12H2O(s)+3H2(g)

Yields of Alum

Mass of aluminum: 1.0006 grams

Mass of dry alum: 12.354 grams

Theoretical Yield:

1.0006/26.98 = .03709 moles Al (Mass Aluminum/Molar Mass)

.03709 * 474.3884g/mole Alum = 17.595 grams Alum (Moles Al/Molar Mass of Alum)

Percent Yield

Actual/Theoritical x 100%

12.3549/17.595 = .7022 *100= 70.22%

Answer 1

Ans:

For the given reaction

2Al(s)+2KOH(aq)+4H2SO4(aq)+22H2O(l)=2KAl(SO4)2*12H2O(s)+3H2(g)

(I) we expect to get 2*26.98g = 53.96g of Al in the form of Alum  if we use 2mols of Al as reactant.

Now we have taken 1.0006g of Al as reactant = (1.0006   26.98) mole = 0.03709 mole of Al.

As the number of moles of the Aluminium taken and the Aluminium present in the Alum as product are same. Therefore the theoretical yield of Al in the form of alum will also be the same and is = 0.03709 mole of Al

(II) From the given equation we can see that the the number of moles of the Alum produced is same with the number of moles of Al used as starting material.

Now the molecular weight of the Alum is =  474.3884g/mole.

and the amount of Al taken as starting material is = 0.03709 mole.

**Hence the theoretical yield of Alum is = No. of moles of Al taken Molecular weight of Alum

= (0.03709 mole 474.3884 g/mole) =17.5951 g of Alum.

(III) Now according to the experimental data we get (68.8550 - 56.5001) g = 12.3549 g of Alum.

**Therefore the % of yield of Alum is = (Experimental weight of Alum Theoretical weight of Alum) 100% = (12.3549 17.5951) 100%

= 70.22%.