To synthesize alum, KAl(SO 4 ) 2 • 12H 2 O, 1.39 g of solid aluminum are reacted

Question

To synthesize alum, KAl(SO4)2 • 12H2O,  1.39 g of solid aluminum are reacted with excess potassium hydroxide and sulfuric acid as described in the experimental section of the writeup in the manual.

1. How many moles of aluminum were obtained?

2. How many moles of KAl(SO4)2 • 12H2O could be formed from the amount of aluminum given?

3. With the amount of aluminum given, what is the theoretical yield of KAl(SO4)2 • 12H2O in grams?

4. After filtering, the amount of KAl(SO4)2 • 12H2O synthisized is found to be  16.67 g . What is the percent yield of KAl(SO4)2 • 12H2O?

This is my prelab. Greatly appreciated if you could show work so I know how to do it for my actual lab. Thanks a ton!!!

Explanation / Answer

KAl(SO4)2 • 12H2O + KOH + H2SO4

balance:

2 Al + 2 KOH + 2 H2SO4 + 22 H2O = 2 KAl(SO4)*12H2O + H2

Q1.

moles of aluminium present:

mol of Al = mass/MW = 1.39/26.98 = 0.05151 mol of Al

Q2.

theoretically,

1 mol of Alum requires = 2 mol of Al

x mol of Alum = 0.05151 mol of Al

x = 0.05151 /2 = 0.025755 mol of alum will be formed

Q3

find yield

mass = mol*MW = 0.025755*474.3884 = 12.2178 g of alum theoretically expected

Q4

if poruct = 16.67 g

There must have been an error,

since teoretial is 12.2 g , and this is much more

there could have been the next error/malpractices:

- KOH is not well diluted, remained as solid

- there are side reaction

- balance is not well calibrated/tared

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