what mass of aluminim would be completely oxidized by 44.8L of oxygen to produce aluminium oxide at stp
At stp, that is 2 moles of O2
2Al + 3O2 >>> Al2O3, so it will take 2/3 of 2 moles Al, or 1.333 moles Al. convert that to grams.
72g
Well, I'd say it's a gas to know. But don't aluminum about it, I'm ready to lend you a helping hand.
To solve this problem, we need to use stoichiometry. But before we dive in, let me just say that I'm pretty good at handling stoichiometry – it's like my second language. Let's go!
The balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide (Al2O3) is:
4 Al + 3 O2 → 2 Al2O3
From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
Now, we have 44.8 L of oxygen at STP. In order to determine the number of moles of oxygen, we can use the ideal gas law. At STP, 1 mole of any ideal gas occupies 22.4 L of space.
So, 44.8 L of oxygen at STP is equal to 44.8/22.4 = 2 moles of oxygen.
According to the balanced equation, 3 moles of oxygen are required to react with 4 moles of aluminum. Therefore, since we have 2 moles of oxygen, we can calculate the moles of aluminum required as follows:
(2 mol O2) × (4 mol Al / 3 mol O2) = 8/3 ≈ 2.67 mol Al
Finally, to determine the mass of aluminum, we need to know its molar mass. Aluminum has a molar mass of approximately 27 g/mol. So, we can calculate the mass using the equation:
mass = moles × molar mass
mass = 2.67 mol × 27 g/mol
mass ≈ 72 g
Hence, approximately 72 grams of aluminum would be completely oxidized by 44.8 liters of oxygen to produce aluminum oxide at STP.
To determine the mass of aluminum that would be completely oxidized by 44.8 liters of oxygen to produce aluminum oxide at STP (Standard Temperature and Pressure), you will need to follow these steps:
Step 1: Write and balance the chemical equation
Write the balanced equation for the reaction between aluminum (Al) and oxygen (O2) to produce aluminum oxide (Al2O3):
4 Al + 3 O2 → 2 Al2O3
Step 2: Convert volume of oxygen to moles
Convert the given volume of oxygen, 44.8 liters, to moles using the ideal gas law equation: PV = nRT. At STP, the pressure (P) is 1 atmosphere (atm), the volume (V) is 22.4 liters, the number of moles (n) can be calculated.
n = PV / RT
Substitute the values:
n = (1 atm) * (44.8 L) / (0.0821 atm L/mol K * 273 K)
n = 1.95 moles
Step 3: Determine the mole ratio
Use the balanced chemical equation from Step 1 to determine the mole ratio between aluminum and oxygen. According to the equation, 4 moles of aluminum react with 3 moles of oxygen.
Step 4: Calculate the moles of aluminum
With the mole ratio from Step 3, you can calculate the moles of aluminum:
n(Al) = (1.95 moles O2) * (4 moles Al / 3 moles O2)
n(Al) = 2.6 moles
Step 5: Convert moles of aluminum to mass
To convert moles of aluminum to mass, you need to know the molar mass of aluminum, which is 26.98 g/mol.
Mass = n * molar mass
Mass(Al) = 2.6 moles * 26.98 g/mol
Mass(Al) ≈ 70.04 g
Therefore, approximately 70.04 grams of aluminum would be completely oxidized by 44.8 liters of oxygen to produce aluminum oxide at STP.