_A2(SO3)3+_NaOH->_Na2SO3+_Al(OH)3

if you start with 389.4 g of Al2(SO3)3 and you produce 212.4 g of Na2SO3, what is your percent yield for this reaction?

First thing you always do; balance the equation.

Convert 389.4 g Al2(SO4)3 to mols. mol = grams/molar mass.
User the coefficients in the balanced equation to convert mols Al2(SO4)3 to mols Na2SO3.
Convert mols Na2SO3 to grams. g = mols x molar mass. This is the theoretical yield (TY). The actual yield(AY) listed in the problem is 212.4.
%yield = (AY/TY)*100 = ?

Well, let's put on our chemistry clown nose and calculate the percent yield for this reaction!

To find the percent yield, we need to use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) x 100

First, let's find the theoretical yield, which is the maximum amount of Na2SO3 that could be produced from the given amount of Al2(SO3)3.

The molar mass of Al2(SO3)3 is:
2(26.98 g/mol) + 3(32.07 g/mol + 3(16.00 g/mol)) = 342.26 g/mol

Now, let's calculate the theoretical yield:

Theoretical Yield = (Mass of Al2(SO3)3 / Molar mass of Al2(SO3)3) x Molar ratio x Molar mass of Na2SO3

Mass of Al2(SO3)3 = 389.4 g
Molar ratio = 2 (from the balanced equation)
Molar mass of Na2SO3 = 81.07 g/mol

So, Theoretical Yield = (389.4 g / 342.26 g/mol) x 2 x 81.07 g/mol

Actual Yield = 212.4 g

Now let's calculate the percent yield:

Percent Yield = (212.4 g / [(389.4 g / 342.26 g/mol) x 2 x 81.07 g/mol]) x 100

= (212.4 g / 381.83 g/mol) x 100

And the answer is...

Drumroll, please!

*Clown drum noises*

The percent yield for this reaction is approximately 55.63%.

Remember, this is just an estimate, so don't juggle the numbers too seriously!

To calculate the percent yield for a reaction, you need to compare the actual yield (the mass of the product actually obtained) with the theoretical yield (the maximum possible amount of product that could be obtained). The percent yield can be calculated using the following formula:

Percent Yield = (Actual Yield / Theoretical Yield) x 100

In this case, the actual yield is given as 212.4 g of Na2SO3. To find the theoretical yield, we need to calculate the molar mass of Al2(SO3)3 and Na2SO3.

The molar mass of Al2(SO3)3:
Al = 26.98 g/mol
S = 32.06 g/mol
O = 16.00 g/mol

Molar mass of Al2(SO3)3 = (2 x 26.98 g/mol) + (3 x (32.06 g/mol + 3 x 16.00 g/mol))
= 2 x 26.98 g/mol + 3 x (32.06 g/mol + 48.00 g/mol)
= 2 x 26.98 g/mol + 3 x 80.06 g/mol
= 53.96 g/mol + 240.18 g/mol
= 294.14 g/mol

Now, we can calculate the theoretical yield of Na2SO3 using stoichiometry. The balanced equation for the reaction is:

2 Al2(SO3)3 + 6 NaOH -> 3 Na2SO3 + 4 Al(OH)3

From the equation, we can see that the molar ratio between Al2(SO3)3 and Na2SO3 is 2:3. Therefore, we calculate the moles of Na2SO3:

Moles of Na2SO3 = (moles of Al2(SO3)3) x (3 moles of Na2SO3 / 2 moles of Al2(SO3)3)

To find the moles of Al2(SO3)3, we use its molar mass and the mass given (389.4 g):

Moles of Al2(SO3)3 = (mass of Al2(SO3)3) / (molar mass of Al2(SO3)3)
= 389.4 g / 294.14 g/mol
= 1.3224 mol

Now we can calculate the moles of Na2SO3:

Moles of Na2SO3 = (1.3224 mol) x (3 mol of Na2SO3 / 2 mol of Al2(SO3)3)
= 1.9836 mol

To find the theoretical yield, we multiply the moles of Na2SO3 by its molar mass:

Theoretical Yield = (moles of Na2SO3) x (molar mass of Na2SO3)
= 1.9836 mol x (2 x 22.99 g/mol) + 32.06 g/mol)
= 1.9836 mol x 126.04 g/mol
= 250.27 g

Finally, we can calculate the percent yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100
= (212.4 g / 250.27 g) x 100
= 84.88%

Therefore, the percent yield for this reaction is approximately 84.88%.

To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of product you obtained) to the theoretical yield (the amount of product you should have obtained based on stoichiometric calculations).

1. Determine the molar masses:
- Al2(SO3)3: 2(26.98 g/mol) + 3(32.07 g/mol + 16.00 g/mol) = 342.16 g/mol
- Na2SO3: 2(22.99 g/mol) + 32.07 g/mol + 3(16.00 g/mol) = 126.04 g/mol

2. Convert the given masses to moles:
- Al2(SO3)3: 389.4 g / 342.16 g/mol = 1.14 mol
- Na2SO3: 212.4 g / 126.04 g/mol = 1.68 mol

3. Use the balanced equation to determine the stoichiometric ratio between Al2(SO3)3 and Na2SO3. From the equation, we can see that 1 mol of Al2(SO3)3 produces 1 mol of Na2SO3.

4. Calculate the theoretical yield:
Since the stoichiometric ratio is 1:1, the theoretical yield of Na2SO3 is equal to the amount of Al2(SO3)3 used (in moles). Therefore, the theoretical yield is 1.14 mol.

5. Calculate the percent yield:
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
- Percent yield = (Actual yield / Theoretical yield) * 100
- Percent yield = (1.68 mol / 1.14 mol) * 100 = 147.37%

Therefore, the percent yield for this reaction is approximately 147.37%.