# Finding percent yield of Alum?

1. Sep 24, 2013

### physicsnobrain

1. The problem statement, all variables and given/known data
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

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

3. The attempt at a solution

First I find the LR. Which in this expirment 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 Im wrong because this is impossible.

What did I do wrong?

2. Sep 24, 2013

### Staff: Mentor

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.

3. Sep 24, 2013

### physicsnobrain

Well we also recieved 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?

4. Sep 24, 2013

### physicsnobrain

Hmm I still get a yield more than 100% after doing this. Which is physically impossible

5. Sep 24, 2013

### physicsnobrain

Is it allowed if I write the formula form the first equation except replace KAl(OH) for KAl(SO4)2

6. Sep 24, 2013

### physicsnobrain

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.

7. Sep 24, 2013

### Saitama

The formula is still wrong. Search for it on google.

8. Sep 24, 2013

### physicsnobrain

I get 61.1% yield. is this correct?

9. Sep 25, 2013

### physicsnobrain

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. Sep 25, 2013

### Staff: Mentor

That looks OK to me.