Consider the following reaction.
4Fe + 3O2 2Fe2O3
a. How many moles of O2 are required to react completely with 46.7 mg of Fe?
How many grams Fe2O3, are produced from the complete reaction of 20.1 g of Fe?
You need to find the arrow key.
4Fe + 3O2 ==> 2Fe2O3
a.
mols Fe = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Fe to mols O2.
b.
mols Fe = grams/atomic mass = ?
Using the coefficients in the balanced equation convert mols Fe to mols Fe2O3.
Then convert mols Fe2O3 to grma. g = mols x molar mass = ?
To determine the number of moles of O2 required to react completely with 46.7 mg of Fe, we first need to convert the mass of Fe to moles.
1. Calculate the molar mass of Fe:
The molar mass of Fe is 55.845 g/mol.
2. Convert the mass of Fe to moles:
(46.7 mg) / (1000 mg/g) = 0.0467 g
0.0467 g / 55.845 g/mol = 0.000836 mol Fe
The balanced chemical equation tells us that the stoichiometric ratio between Fe and O2 is 4:3. Therefore, for every 4 moles of Fe, we need 3 moles of O2.
3. Calculate the number of moles of O2 required:
(0.000836 mol Fe) * (3 mol O2 / 4 mol Fe) = 0.000627 mol O2
So, 0.000627 moles of O2 are required to react completely with 46.7 mg of Fe.
Now, let's calculate the grams of Fe2O3 produced from the complete reaction of 20.1 g of Fe.
4. Convert the mass of Fe to moles:
(20.1 g) / (55.845 g/mol) = 0.359 mol Fe
Again, using the stoichiometric ratio from the balanced equation, we know that for every 4 moles of Fe, we get 2 moles of Fe2O3.
5. Calculate the number of moles of Fe2O3:
(0.359 mol Fe) * (2 mol Fe2O3 / 4 mol Fe) = 0.18 mol Fe2O3
Lastly, we need to convert the moles of Fe2O3 to grams.
6. Calculate the mass of Fe2O3:
0.18 mol Fe2O3 * (159.69 g/mol) = 28.74 g Fe2O3
Therefore, 28.74 grams of Fe2O3 are produced from the complete reaction of 20.1 grams of Fe.
To determine the number of moles of O2 required to react with 46.7 mg of Fe, we can use stoichiometry and the given balanced equation.
1. Start by converting the mass of Fe to moles. We can use the molar mass of Fe to do this conversion.
Molar mass of Fe = 55.85 g/mol
Moles of Fe = (mass of Fe / molar mass of Fe)
= (46.7 mg / 1000 mg/g) / 55.85 g/mol
= 0.000835 mol
2. Use stoichiometry to determine the moles of O2 required.
According to the balanced equation, the molar ratio of Fe to O2 is 4:3.
Moles of O2 = (moles of Fe) * (3 moles of O2 / 4 moles of Fe)
= 0.000835 mol * (3/4)
= 0.000626 mol
Therefore, 0.000626 moles of O2 are required to react completely with 46.7 mg of Fe.
To determine the grams of Fe2O3 produced from the complete reaction of 20.1 g of Fe, we'll again use stoichiometry.
1. Convert the mass of Fe to moles using the molar mass of Fe.
Moles of Fe = 20.1 g / 55.85 g/mol
= 0.359 mol
2. Use stoichiometry to determine the moles of Fe2O3 produced.
According to the balanced equation, the molar ratio of Fe to Fe2O3 is 4:2.
Moles of Fe2O3 = (moles of Fe) * (2 moles of Fe2O3 / 4 moles of Fe)
= 0.359 mol * (2/4)
= 0.1795 mol
3. Convert moles of Fe2O3 to grams using the molar mass of Fe2O3.
Molar mass of Fe2O3 = 159.69 g/mol
Grams of Fe2O3 = (moles of Fe2O3) * (molar mass of Fe2O3)
= 0.1795 mol * 159.69 g/mol
= 28.68 g
Therefore, 28.68 grams of Fe2O3 are produced from the complete reaction of 20.1 g of Fe.