What Did I Do Wrong in Finding the Percent Yield of Alum?

AI Thread Summary
The discussion revolves around calculating the percent yield of alum from an experiment where 0.766g of aluminum produced 8.246g of alum. The initial calculations were based on an incorrect formula for alum, KAl(OH)4, leading to an impossible yield of 217%. After revising the formula to KAl(SO4)2 * 12H2O, a more reasonable yield of 61.1% was obtained. Participants emphasized the importance of using the correct chemical formula to avoid erroneous results. The final consensus suggests that the revised formula yields a plausible percent yield.
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Homework Statement


Find the percent yield of alum

In an experiment to synthesize alum I started with 0.766g of aluminum and made 8.246g of alum


Homework Equations


2Al + 6H20 + 2KOH ---> 2KAl(OH)4 + 3H2


The Attempt at a Solution



First I find the LR. Which in this experiment is Al.

I find the moles of aluminum to be 0.02838mol, next I multiply by the mole ratio and by the molar mass of alum to find the theoretical yield of alum.

Now to find the percent yield I do actual/theoretical X100%

8.246g/3.806g x 100% = 217%. I know I am wrong because this is impossible.

What did I do wrong?
 
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Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.
 
Borek said:
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.

Well we also received another formula in this lab:

KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

Should I add this to the formula I listed in my first post to get a overall reaction equation?
 
physicsnobrain said:
Well we also received another formula in this lab:

KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

Should I add this to the formula I listed in my first post to get a overall reaction equation?

Hmm I still get a yield more than 100% after doing this. Which is physically impossible
 
Borek said:
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.


Is it allowed if I write the formula form the first equation except replace KAl(OH) for KAl(SO4)2
 
Borek said:
Your calculations are OK, but they are based on the incorrect alum formula - it is not KAl(OH)4. Unless you just use a wrong name of the compound.

Also, even if I use the formula u refer to I get a percent yield of 112% which is physically impossible in every sense and wrong.
 
physicsnobrain said:
Well we also received another formula in this lab:

KAl(OH)4 +8H2O +2H2SO4 ---> KAl(SO4)2 * 12H20

Should I add this to the formula I listed in my first post to get a overall reaction equation?

The formula is still wrong. Search for it on google.
 
Pranav-Arora said:
The formula is still wrong. Search for it on google.

I get 61.1% yield. is this correct?
 
physicsnobrain said:
I get 61.1% yield. is this correct?

I got this number using the fomula:

2Al +2KOH +4H2SO4 + 22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O


This was the only formula that helped produce a reasonable percent yield (61.1%). I am highly sceptical that there is an alternative to this.
 
  • #10
That looks OK to me.
 
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