# Finding percent yield of Alum?

• physicsnobrain
In summary, to find the percent yield of alum, the correct formula for the reaction is 2Al +2KOH +4H2SO4 + 22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O. Using this formula, the calculated yield was 61.1%, which is a reasonable and possible result. Other formulas given during the experiment were incorrect and would result in a physically impossible yield.

## 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?

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.

That looks OK to me.

## 1. What is the purpose of finding the percent yield of Alum?

The purpose of finding the percent yield of Alum is to determine the efficiency of a chemical reaction. It compares the actual amount of Alum produced in a reaction to the theoretical amount that should have been produced based on the reactants used. This information can help scientists improve reaction conditions and optimize their processes.

## 2. How do you calculate the percent yield of Alum?

The percent yield of Alum is calculated by dividing the actual yield (the amount of Alum produced in the reaction) by the theoretical yield (the amount of Alum that should have been produced based on the reactants used) and multiplying by 100%. The formula is: (actual yield/theoretical yield) x 100%.

## 3. What factors can affect the percent yield of Alum?

There are several factors that can affect the percent yield of Alum, including impurities in the reactants, incomplete reactions, and side reactions that produce unwanted products. Additionally, experimental errors such as inaccurate measurements or loss of product during handling can also impact the yield.

## 4. What is considered a good percent yield of Alum?

A good percent yield of Alum will vary depending on the specific reaction and experimental conditions. However, in general, a percent yield of 80-90% is considered good. This means that the reaction is efficient and most of the reactants are being converted to Alum.

## 5. How can the percent yield of Alum be improved?

To improve the percent yield of Alum, scientists can try to optimize reaction conditions such as temperature, concentration, and reaction time. They can also purify the reactants to remove impurities and minimize side reactions. Accurate measurements and careful handling of the product can also help improve the yield.