Calculate enthalpy change when 35.0 grams barium oxide reacts with 5.00 grams of aluminum.
0.228 mol BaO ( limiting reagent)
0.1853 mol Al
Molar Ratio 1.23
-1669.8 - 3(-558.1) = 3344.1 kJ/mol
0.228 BaO reacts therefore:
(0.228/3) x 3344.1 = 254.2 kJ
I can't confirm those dH formation values but I'll assume you have them right. In calculating, however, when you finish isn't that 3344.1 kJ that many kJ for the reaction as shown and not kJ/mol.
It appears you used dHrxn = (n*dHformation products) - (n*dH formation reactants). Having said all of that, however, you did not multiply by 3 and you used the 3344.1 correctly with the 0.228. Other than that I don't see anything wrong with this.
Well, calculating the enthalpy change can be a bit dry, but let's try to add some humor to it.
So, imagine Barium Oxide and Aluminum walking into a bar. Barium Oxide is feeling pretty confident, strutting his stuff because he's got more weight on him. But little does he know that Aluminum has some tricks up his sleeve.
They start a conversation, and Aluminum says, "Hey Barium Oxide, I challenge you to a reaction! Let's see who's got more energy!"
Barium Oxide, being all cocky, says, "Alright, bring it on!"
And so, the reaction begins! They mix together and go through a wild dance. But little did Barium Oxide know, Aluminum was the one calling all the shots. He was the "limiting reagent" in the reaction, meaning he controlled the show.
After the reaction, they both take a step back, exhausted from all the dancing. Barium Oxide looks disappointed, realizing that Aluminum had stolen the show.
Now, let's get to the boring part. Through some calculations, we find that the enthalpy change, or the energy released during the reaction, is approximately 254.2 kJ. So Aluminum really showed Barium Oxide who's the boss!
But hey, at least Barium Oxide gave it his best shot, right? Just remember, chemistry can be a bit serious, but adding some humor makes it more entertaining!
To calculate the enthalpy change when 35.0 grams of barium oxide reacts with 5.00 grams of aluminum, we first need to determine the limiting reagent.
First, we calculate the number of moles of each compound:
Molar mass of BaO = 137.33 g/mol
Molar mass of Al = 26.98 g/mol
Number of moles of BaO = mass / molar mass = 35.0 g / 137.33 g/mol = 0.255 mol
Number of moles of Al = mass / molar mass = 5.00 g / 26.98 g/mol = 0.1853 mol
Next, we determine the molar ratio of the reaction:
BaO + 3Al -> Ba + Al2O3
From the balanced equation, we see that for every 1 mole of BaO, we need 3 moles of Al.
Molar ratio = (moles of Al) / (moles of BaO) = 0.1853 mol / 0.255 mol ≈ 0.725
Since the molar ratio is less than 1, Al is the limiting reagent, and BaO is in excess.
Next, we calculate the enthalpy change for the reaction:
ΔH = ΔH_rxn = ΔH_products - ΔH_reactants
From the given data:
ΔH_rxn = -1669.8 kJ/mol - 3(-558.1 kJ/mol) = 3344.1 kJ/mol
Since 0.228 moles of BaO reacts, we determine the enthalpy change for this amount:
ΔH = (moles of BaO reacted) x (ΔH_rxn) = (0.228/3) x 3344.1 kJ/mol ≈ 254.2 kJ
Therefore, the enthalpy change for the reaction when 35.0 grams of BaO reacts with 5.00 grams of Al is approximately 254.2 kJ.
To calculate the enthalpy change when barium oxide reacts with aluminum, we need to determine the limiting reagent, molar ratios, and the enthalpy change per mole of reactant.
Step 1: Determine the limiting reagent
To find the limiting reagent, we compare the number of moles of barium oxide and aluminum. The molar masses of BaO and Al are:
- BaO: 137.33 g/mol
- Al: 26.98 g/mol
Converting the given masses to moles:
- Moles of BaO = 35.0 g / 137.33 g/mol = 0.2551 mol
- Moles of Al = 5.00 g / 26.98 g/mol = 0.1853 mol
Since the molar ratio of BaO to Al is 1:3, we can see that there is an excess amount of BaO. Therefore, BaO is the limiting reagent.
Step 2: Calculate the molar ratio
The molar ratio of BaO to Al is given as 1.23 (which is approximately 1:1). This means that for every 1 mole of BaO, 1.23 moles of Al are needed.
Step 3: Determine the enthalpy change per mole
The enthalpy change per mole is given as -558.1 kJ/mol for aluminum. However, we need to cancel out the stoichiometric coefficient in the balanced equation to match the molar ratio of BaO and Al.
The balanced equation is:
3 BaO + 2 Al -> 3 Ba + Al2O3
Comparing the coefficients, we can see that 1 mole of Al has a coefficient of 2, so we need to multiply the enthalpy change per mole by 2:
-558.1 kJ/mol * 2 = -1116.2 kJ/mol
Step 4: Calculate the overall enthalpy change
Since we know that the molar ratio of BaO to Al is 1.23 and the enthalpy change per mole for Al is -1116.2 kJ/mol, we can calculate the overall enthalpy change using the formula:
Overall enthalpy change = (molar ratio of BaO/3) * enthalpy change per mole of Al
Substituting the values:
Overall enthalpy change = (0.228/3) * (-1116.2 kJ/mol) = -84.2 kJ/mol
Therefore, the enthalpy change for the reaction is -84.2 kJ/mol.