Calculate the theoretical yield of alum if 0.200g of Aluminum was reacted with excess KOH and H2SO4?
I don't even know how to start it. Please help. Also I wanted to say thank you for the help I've gotten from this website it has helped me a lot. I feel like I'm understanding the material better ever since I found it.
You're welcome! I'm glad to hear that the website has been helpful for you. I'll be happy to guide you through the calculation for the theoretical yield of alum.
To calculate the theoretical yield, we need to determine the balanced chemical equation for the reaction of aluminum with KOH and H2SO4. The reaction can be represented as follows:
2Al + 2KOH + 4H2SO4 -> K2SO4 + Al2(SO4)3 + 6H2O
From the balanced equation, we can see that 2 moles of aluminum (Al) react with 2 moles of KOH and 4 moles of H2SO4 to produce 1 mole of alum, Al2(SO4)3.
The molar mass of aluminum (Al) is 26.98 g/mol and the molar mass of alum (Al2(SO4)3) is 342.15 g/mol.
First, we need to calculate the moles of aluminum used:
moles Al = mass / molar mass of Al
moles Al = 0.200 g / 26.98 g/mol = 0.00741 mol
Since aluminum is given in excess, it means there is more than enough aluminum to react completely with KOH and H2SO4.
Next, we can calculate the moles of alum produced using the stoichiometry of the balanced equation:
moles Alum = moles Al
moles Alum = 0.00741 mol
Finally, we can calculate the theoretical yield of alum in grams:
theoretical yield of Alum = moles Alum × molar mass of alum
theoretical yield of Alum = 0.00741 mol × 342.15 g/mol = 2.53 g
Therefore, the theoretical yield of alum is 2.53 grams, assuming all reactants are completely consumed and the reaction goes to completion.
You're welcome! I'm glad to hear that this website has been helpful to you. I'm here to assist you with your question on calculating the theoretical yield of alum.
To calculate the theoretical yield, you need to follow a few steps:
Step 1: Write and balance the chemical equation.
You already mentioned that aluminum (Al) reacts with excess KOH (potassium hydroxide) and H2SO4 (sulfuric acid) to form alum. The balanced chemical equation for this reaction is:
2Al + 2KOH + 4H2SO4 → 2KAl(SO4)2 · 12H2O + 3H2
Step 2: Convert the given mass to moles.
In this case, you're given that the mass of aluminum (Al) is 0.200g. You need to convert this mass to moles by using the molar mass of aluminum. The molar mass of aluminum is 26.98 g/mol.
Dividing the given mass by the molar mass will give you the number of moles:
0.200g Al * (1 mol Al / 26.98g Al) = 0.00741 mol Al
Step 3: Determine the limiting reactant.
To determine the limiting reactant, you need to compare the mole ratios of the given reactants to the balanced chemical equation. Here, you have excess KOH and H2SO4, but you need to determine which reactant limits the reaction.
From the balanced equation, you can see that the mole ratio of Al : KAl(SO4)2 · 12H2O is 2:2. This means that 2 moles of Al will produce 2 moles of KAl(SO4)2 · 12H2O.
Step 4: Calculate the theoretical yield.
Since you have determined that Al is the limiting reactant, you can use the mole ratio from the balanced equation to calculate the theoretical yield of alum.
0.00741 mol Al * (2 mol KAl(SO4)2 · 12H2O / 2 mol Al) = 0.00741 mol KAl(SO4)2 · 12H2O
Finally, you can convert the moles of KAl(SO4)2 · 12H2O to grams using its molar mass. The molar mass of KAl(SO4)2 · 12H2O is 474.38 g/mol.
0.00741 mol KAl(SO4)2 · 12H2O * (474.38g KAl(SO4)2 · 12H2O / 1 mol KAl(SO4)2 · 12H2O) = 3.51 g KAl(SO4)2 · 12H2O
Therefore, the theoretical yield of alum is 3.51 grams. Remember, this is the maximum amount of alum that could be produced assuming 100% efficiency. Actual yields may be lower due to factors such as incomplete reactions or side reactions.
2Al ==> Al2(SO4)3.K2SO4.24H2O
The above isn't fully balanced but I've balanced the only part that matters.
mols Al = grams/molar mass = approx 0.0074
Using the coefficients in the equation convert mols Al to mols of alum. That's 0.0074 x (1 mol alum/2 mol Al) = 0.0074 x 1/2 = 0.0037
g alum produced = mols alum x molar mass alum.
NOTE: Some write the formula for alum as the above; others write it as half that (that is KAl(SO4)2.12H2O but that will not change the answer you get.)