Given the following balanced equation, determine the mass of iron metal that would be produced from the complete reaction of 94.7g of aluminum with excess iron II oxide.
2 Al(s) + 3FeO(aq) → 3Fe(s) + Al2O(aq)
Well, we can start by finding the molar mass of aluminum (Al) and iron (Fe). The molar mass of aluminum is 26.98 g/mol, and the molar mass of iron is 55.85 g/mol.
Next, we want to find the number of moles of aluminum we have. We can use the formula:
moles = mass / molar mass
So, moles of aluminum = 94.7 g / 26.98 g/mol = 3.511 mol
According to the balanced equation, we need 2 moles of aluminum to produce 3 moles of iron.
Using this ratio, we can determine the number of moles of iron produced:
moles of iron = (3 moles of iron / 2 moles of aluminum) * 3.511 moles of aluminum = 5.267 mol
Finally, we can find the mass of iron produced:
mass of iron = moles of iron * molar mass of iron = 5.267 mol * 55.85 g/mol = 294.12 g
So, the mass of iron metal that would be produced is approximately 294.12 grams.
But hey, keep in mind that this is just theoretical. In reality, you might not actually get that exact mass due to various factors like experimental error and the limitations of the reaction. So take it with a pinch of irony!
To determine the mass of iron metal produced, we can use stoichiometry calculations based on the balanced equation.
The balanced equation tells us that 2 moles of aluminum (Al) react with 3 moles of iron II oxide (FeO) to produce 3 moles of iron (Fe).
Step 1: Calculate the number of moles of aluminum (Al) present in 94.7g of aluminum.
To do this, we need to know the molar mass of aluminum (Al), which is 26.98 g/mol.
Number of moles of Al = mass of Al / molar mass of Al
= 94.7g / 26.98 g/mol
Step 2: Calculate the number of moles of iron (Fe) produced using stoichiometry.
From the balanced equation, we know that 2 moles of Al react to produce 3 moles of Fe.
Number of moles of Fe = (number of moles of Al) x (3 moles of Fe / 2 moles of Al)
Step 3: Calculate the mass of iron (Fe) produced.
To do this, we need to know the molar mass of iron (Fe), which is 55.85 g/mol.
Mass of Fe = number of moles of Fe x molar mass of Fe
Now, let's plug in the values:
Number of moles of Al = 94.7g / 26.98 g/mol = 3.51 mol (rounded to two decimal places)
Number of moles of Fe = 3.51 mol x (3 mol Fe / 2 mol Al) = 5.27 mol (rounded to two decimal places)
Mass of Fe = 5.27 mol x 55.85 g/mol = 294.38 g (rounded to two decimal places)
Therefore, the mass of iron metal produced from the complete reaction of 94.7g of aluminum with excess iron II oxide is approximately 294.38 grams.
To determine the mass of iron metal produced, we first need to calculate the amount of aluminum that will react.
1. Convert the given mass of aluminum to moles.
The molar mass of aluminum (Al) is 26.98 g/mol.
Moles of aluminum = Mass of aluminum / Molar mass of aluminum
= 94.7 g / 26.98 g/mol
≈ 3.51 mol (rounded to two decimal places)
2. Use the stoichiometry coefficients from the balanced equation to determine the moles of iron formed.
According to the balanced equation: 2 moles of aluminum react with 3 moles of iron to produce 3 moles of iron.
Moles of iron = (Moles of aluminum) x (3 moles of iron / 2 moles of aluminum)
= 3.51 mol x (3/2)
≈ 5.27 mol (rounded to two decimal places)
3. Calculate the mass of iron.
The molar mass of iron (Fe) is 55.85 g/mol.
Mass of iron = Moles of iron x Molar mass of iron
= 5.27 mol x 55.85 g/mol
≈ 294.66 g (rounded to two decimal places)
Therefore, approximately 294.66 grams of iron metal will be produced from the complete reaction of 94.7 grams of aluminum with excess iron II oxide.
The equation you wrote is not correct. It should be Al2O3(s). Also, I don't know how you make FeO and Al2O3 in aqueous solutions.
2Al + 3FeO → 3Fe + Al2O3
mols Al = g/atomic mass = 94.7g/27 = 3.51. Convert to mols Fe this way.
3.51 mols Al x (3 mols Fe/2 mol Al) = 5.26 mols Fe produced.
Convert to g Fe this way. grams Fe = mols Fe x atomic mass Fe = ?