Al2(SO3)3 + 6NaOH �¨ 3Na2SO3 + 2Al(OH)3
If this reaction begins with 10 g Al2(SO3)3 and 10 g NaOH, what is the theoretical yield in grams Al(OH)3 and what is the limiting reactant?
To find the theoretical yield of Al(OH)3 and the limiting reactant, we need to determine how many moles of each reactant are present and compare their stoichiometric ratios.
Step 1: Convert the given masses of Al2(SO3)3 and NaOH to moles.
The molar mass of Al2(SO3)3 is:
Al = 26.98 g/mol x 2 = 53.96 g/mol
S = 32.07 g/mol x 3 = 96.21 g/mol
O = 16.00 g/mol x 9 = 144.00 g/mol
Total molar mass = 53.96 g/mol + 96.21 g/mol + 144.00 g/mol = 294.17 g/mol
Number of moles of Al2(SO3)3 = Mass / Molar mass
Number of moles of Al2(SO3)3 = 10 g / 294.17 g/mol = 0.034 moles
The molar mass of NaOH is:
Na = 22.99 g/mol
O = 16.00 g/mol + 1.01 g/mol + 1.01 g/mol = 18.02 g/mol
Total molar mass = 22.99 g/mol + 18.02 g/mol = 40.01 g/mol
Number of moles of NaOH = Mass / Molar mass
Number of moles of NaOH = 10 g / 40.01 g/mol = 0.250 moles
Step 2: Calculate the stoichiometric ratios between Al2(SO3)3 and Al(OH)3.
According to the balanced equation, the stoichiometric ratio between Al2(SO3)3 and Al(OH)3 is 2:2 or 1:1.
Step 3: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of Al2(SO3)3 and NaOH based on the stoichiometric ratio.
Moles of Al2(SO3)3: 0.034 moles
Moles of NaOH (multiplied by the stoichiometric ratio): 0.250 moles x (1 mole NaOH / 6 moles NaOH) = 0.042 moles
The moles of Al2(SO3)3 are less than the moles of NaOH, which means Al2(SO3)3 is the limiting reactant.
Step 4: Calculate the theoretical yield of Al(OH)3.
The stoichiometric ratio between Al2(SO3)3 and Al(OH)3 is 2:2 or 1:1.
Theoretical yield of Al(OH)3 (moles) = Moles of Al2(SO3)3 (limiting reactant)
Theoretical yield of Al(OH)3 (moles) = 0.034 moles
To calculate the theoretical yield in grams, we use the molar mass of Al(OH)3:
Molar mass of Al(OH)3 = 26.98 g/mol + 1.01 g/mol + 1.01 g/mol + 1.01 g/mol = 30.11 g/mol
Theoretical yield of Al(OH)3 (grams) = Theoretical yield of Al(OH)3 (moles) x Molar mass of Al(OH)3
Theoretical yield of Al(OH)3 (grams) = 0.034 moles x 30.11 g/mol = 1.023 g
Therefore, the theoretical yield of Al(OH)3 in grams is 1.023 g, and the limiting reactant is Al2(SO3)3.
To find the theoretical yield of Al(OH)3 and identify the limiting reactant, we need to follow these steps:
Step 1: Convert the given masses of Al2(SO3)3 and NaOH to moles.
To do this, we will use the molar masses of Al2(SO3)3 and NaOH, which are:
- Al2(SO3)3: 2(26.98 g/mol) + 3(32.07 g/mol + 3(16.00 g/mol)) = 342.14 g/mol
- NaOH: 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol
The moles of Al2(SO3)3 can be calculated as:
moles of Al2(SO3)3 = mass of Al2(SO3)3 / molar mass of Al2(SO3)3
moles of Al2(SO3)3 = 10 g / 342.14 g/mol ≈ 0.0292 mol
The moles of NaOH can be calculated as:
moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 10 g / 39.99 g/mol ≈ 0.25 mol
Step 2: Determine the stoichiometric ratio between Al2(SO3)3 and Al(OH)3.
From the balanced equation: Al2(SO3)3 + 6NaOH → 3Na2SO3 + 2Al(OH)3
We can see that the ratio between Al2(SO3)3 and Al(OH)3 is 2:2 (or 1:1).
Step 3: Calculate the moles of Al(OH)3 produced.
Since Al2(SO3)3 and Al(OH)3 have a 1:1 stoichiometric ratio, the moles of Al(OH)3 produced will be the same as the moles of Al2(SO3)3 present.
moles of Al(OH)3 = 0.0292 mol (as calculated in Step 1)
Step 4: Calculate the mass of Al(OH)3.
mass of Al(OH)3 = moles of Al(OH)3 * molar mass of Al(OH)3
mass of Al(OH)3 = 0.0292 mol * (2(26.98 g/mol) + 3(16.00 g/mol) + 1.01 g/mol)
mass of Al(OH)3 = 0.0292 mol * 78.00 g/mol ≈ 2.27 g
Therefore, the theoretical yield of Al(OH)3 is approximately 2.27 g.
Step 5: Determine the limiting reactant.
To identify the limiting reactant, we compare the moles of reactant used to the stoichiometry of the balanced equation.
The balanced equation shows that for every 1 mol of Al2(SO3)3, 2 mol of Al(OH)3 are produced.
The moles of Al2(SO3)3 available are 0.0292 mol, and using the stoichiometry, we can expect 0.0292 mol * 2 = 0.0584 mol of Al(OH)3 if all Al2(SO3)3 is consumed.
However, the moles of Al(OH)3 calculated in Step 3 is 0.0292 mol, which is equal to the moles of Al(OH)3 expected from the Al2(SO3)3 used. This suggests that Al2(SO3)3 is the limiting reactant since it is completely consumed, and only 0.0292 mol of Al(OH)3 can be produced.
Hence, the limiting reactant is Al2(SO3)3.
3.44
Al2(SO4)3 + 6NaOH ==>3Na2SO4 + 2Al(OH)3
mols Al2(SO4)3 = grams/molar mass = ?
mols NaOH = grams/molar mass
Using the coefficients in the balanced equation, convert mols Al2(SO4)3 to mols Al(OH)3.
Do the same and convert mols NaOH to mols Al(OH)3.
It is likely that the two values will not agree which means one is not right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that number is the limiting reagent.
Using the smaller number, convert to grams. grams Al(OH)3 = mols Al(OH)3 x molar mass Al(OH)3. That is the theoretical yield .