K+(aq) + Al3+(aq) + 2SO42-(aq) + 2H2O(l)→KAl (SO4)2 * 12H2O(s)
Al is limiting and 1.007 g
Find the theoretical yield and percent yield
I think the answer is 17 and 40, respect.
I just need help figuring it out
Is that 1.007 g Al or 1.007 something different? Is that 12H2O on the left or 2 H2O? I assume 12.
mols Al = 1.007/atomic mass Al
Use the coefficients to change mols Al to mol KAl(SO4)2.12H2O
Now convert to grams. g = mols x molar mass. This is the theoretical yield (TY) and I obtained 17.7 g if your talking KAl(SO4)2.12H2O
%yield = (actual yield/TY)*100 = ? You don't list an actual yield in the problem.
To find the theoretical yield and percent yield, we first need to calculate the moles of Al present in the reaction and then use the stoichiometry to determine the moles of the product formed.
Given:
Mass of Al = 1.007 g
Step 1: Calculate the moles of Al
Molar mass of Al = 26.98 g/mol
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 1.007 g / 26.98 g/mol
Moles of Al = 0.0374 mol
Step 2: Use the stoichiometry to determine the moles of the product formed
From the balanced equation, we can see that the stoichiometric ratio between Al and the product KAl(SO4)2 * 12H2O is 1:1.
Therefore, the moles of the product formed will be equal to the moles of Al.
Moles of the product formed = Moles of Al = 0.0374 mol
Step 3: Calculate the molar mass of the product
Molar mass of KAl(SO4)2 * 12H2O = (39.1 g/mol + 26.98 g/mol + 96.06 g/mol + (4 x 16.00 g/mol) + 2(1.01 g/mol) + 12(1.008 g/mol)) x 2 + 12(1.008 g/mol)
Molar mass of KAl(SO4)2 * 12H2O = 474.40 g/mol
Step 4: Calculate the theoretical yield
Theoretical yield = Moles of the product formed x Molar mass of the product
Theoretical yield = 0.0374 mol x 474.40 g/mol
Theoretical yield = 17.75 g (rounded to two decimal places)
Step 5: Calculate the percent yield
Percent yield = (Actual yield / Theoretical yield) x 100
Given that the mass of Al used is 1.007 g and the theoretical yield is 17.75 g:
Percent yield = (1.007 g / 17.75 g) x 100
Percent yield ≈ 5.68% (rounded to two decimal places)
So, the correct values for the theoretical yield and percent yield are 17.75 g and 5.68%, respectively.
To find the theoretical yield and percent yield, we need to follow a few steps.
Step 1: Write the balanced equation for the reaction.
The balanced equation is:
2Al + 3K2SO4 + 22H2O → 3KAl(SO4)2 * 12H2O
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that 2 moles of aluminum (Al) react with 3 moles of potassium sulfate (K2SO4) to produce 3 moles of potassium aluminum sulfate (KAl(SO4)2 * 12H2O).
Therefore, the stoichiometric ratio between Al and KAl(SO4)2 * 12H2O is 2:3.
Step 3: Calculate the moles of Al given.
The given mass of Al is 1.007 g. To calculate the moles of Al, we need to divide the given mass by the molar mass of Al, which is 26.98 g/mol.
moles of Al = mass of Al / molar mass of Al
moles of Al = 1.007 g / 26.98 g/mol
moles of Al ≈ 0.0374 mol
Step 4: Determine the theoretical yield.
Using the stoichiometric ratio from Step 2, we can determine the theoretical yield of KAl(SO4)2 * 12H2O.
theoretical yield of KAl(SO4)2 * 12H2O = moles of Al x (3 moles of KAl(SO4)2 * 12H2O / 2 moles of Al)
theoretical yield of KAl(SO4)2 * 12H2O = 0.0374 mol x (3 / 2)
theoretical yield of KAl(SO4)2 * 12H2O ≈ 0.0561 mol
Step 5: Convert the theoretical yield from moles to grams.
To calculate the theoretical yield in grams, we need to multiply the moles by the molar mass of KAl(SO4)2 * 12H2O, which is 474.39 g/mol.
theoretical yield = theoretical yield in moles x molar mass of KAl(SO4)2 * 12H2O
theoretical yield = 0.0561 mol x 474.39 g/mol
theoretical yield ≈ 26.61 g
Step 6: Calculate the percent yield.
The percent yield is calculated by comparing the actual yield (given) with the theoretical yield, then dividing by the theoretical yield, and multiplying by 100.
percent yield = (actual yield / theoretical yield) x 100
In this case, the given mass of Al is the actual yield.
percent yield = (1.007 g / 26.61 g) x 100
percent yield ≈ 3.79%
Therefore, the theoretical yield is approximately 26.61 g, and the percent yield is approximately 3.79%.