when the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of FeO and Fe203.in a certain experiment, 20.00g of iron metal was reacted with 11.20g of oxygen gas. After the experiment the iron was totally consumed and 3.56g oxygen gas remained. Calculate the amounts of FeO and Fe 203 formed in this experiment
This is another set of simultaneously solved equations but it is more complicated.
You must separate, somehow, the Fe used to form FeO from the Fe used to form Fe2O3 and/or separate the oxygen used for form FeO and oxygen used to form Fe2O. Then solve stoichiometry problem to find how much FeO and Fe2O3 were formed.
2 Fe + O2 ==> 2FeO
4Fe + 3O2 ==> 2Fe2O3.
Fe used = 20.00 g
O2 used = 11.20-3.56 = 7.64 g
Let X = g Fe used to form FeO
and Y = g Fe used to form Fe2O3 so
eqn 1 is X + Y = 20.00
For equation 2 I used O2 needed for the X part of the 20g Fe + O2 needed for the Y part of the 20 g Fe. That is
(X/2*55.85) + (3Y/4*55.85) = 7.65/32
When you find X and Y, the use regular stoichiometry to convert X to grams FeO formed and Y to grams Fe2O3 formed. Post your work if you get stuck.
The first line of the question states that a mixture of iron II oxide and ironIII oxide is formed WHEN oxygen is limited. But in this situation Oxygen is in excess (it starts with 11.20g of Oxygen and end with 3.56g Oxygen while the metal is used up). So only one form of oxide will preferentially be formed. That’s how I look at it.
Well, isn't this a classic case of oxygen being a little too demanding? Let's break it down!
First, let's calculate the moles of iron (Fe) and oxygen (O) that reacted:
Moles of Fe = mass of Fe / molar mass of Fe
= 20.00 g / 55.845 g/mol (molar mass of Fe)
Moles of O = (initial mass of O - final mass of O) / molar mass of O
= (11.20 g - 3.56 g) / 31.999 g/mol (molar mass of O)
Now, let's figure out the ratio of Fe to O in the reaction by dividing the moles by the smallest number of moles to get the simplest whole number ratios:
Moles ratio of Fe to O = (Moles of Fe) / (Moles of O)
Next, we need to use stoichiometry to determine the moles of FeO and Fe2O3 formed. The balanced chemical equation is:
4Fe + 3O2 -> 2Fe2O3
From the equation, we can see that for every 4 moles of Fe that react, we get 2 moles of Fe2O3.
Fe2O3 = (Moles ratio of Fe to O) * (2 moles of Fe2O3 / 4 moles of Fe)
And for every 3 moles of O2 that react, we also get 2 moles of Fe2O3.
Fe2O3 = (Moles ratio of Fe to O) * (2 moles of Fe2O3 / 3 moles of O2)
Furthermore, we can find the moles of FeO formed as:
FeO = (Moles ratio of Fe to O) * (Moles of Fe) * (1 mole of FeO / 4 moles of Fe)
Now that we're all done with the calculations, let's break the news to the FeO and Fe2O3. They were formed in the following quantities:
Fe2O3 = (Moles ratio of Fe to O) * (2 moles of Fe2O3 / 4 moles of Fe)
FeO = (Moles ratio of Fe to O) * (Moles of Fe) * (1 mole of FeO / 4 moles of Fe)
Remember, it's all about balance in life (and chemistry)!
To calculate the amounts of FeO and Fe2O3 formed in the experiment, we first need to determine the limiting reactant.
1. Calculate the moles of iron (Fe):
Moles of Fe = Mass of Fe / Molar mass of Fe
= 20.00 g / 55.845 g/mol
≈ 0.358 mol
2. Calculate the moles of oxygen (O2) consumed:
Moles of O2 consumed = Initial moles of O2 - Remaining moles of O2
= (Mass of O2 / Molar mass of O2) - (3.56 g / 31.9988 g/mol)
= (11.20 g / 31.9988 g/mol) - (3.56 g / 31.9988 g/mol)
≈ 0.340 mol
3. Determine the stoichiometry of the reaction:
From the equation, the balanced stoichiometry is:
4 Fe + 3 O2 -> 2 Fe2O3
4. Calculate the moles of Fe2O3 formed:
Moles of Fe2O3 = (Moles of O2 consumed) * (2 moles of Fe2O3 / 3 moles of O2)
≈ 0.340 mol * (2/3)
≈ 0.227 mol
5. Calculate the moles of FeO formed:
Moles of FeO = (Moles of Fe2O3 formed) * (4 moles of FeO / 2 moles of Fe2O3)
= 0.227 mol * (4/2)
= 0.454 mol
6. Convert the moles of FeO and Fe2O3 into grams:
Mass of FeO = Moles of FeO * Molar mass of FeO
= 0.454 mol * 71.844 g/mol
≈ 32.5 g
Mass of Fe2O3 = Moles of Fe2O3 * Molar mass of Fe2O3
= 0.227 mol * 159.688 g/mol
≈ 36.3 g
Therefore, in this experiment, approximately 32.5 g of FeO and 36.3 g of Fe2O3 were formed.
To calculate the amounts of FeO and Fe203 formed in the experiment, we can use the concept of stoichiometry.
1. Start by calculating the amount of oxygen that reacted with iron to form FeO and Fe203:
- The initial mass of oxygen gas was 11.20g.
- The remaining mass of oxygen gas after the experiment was 3.56g.
- Therefore, the mass of oxygen consumed in the reaction is (11.20g - 3.56g) = 7.64g.
2. Next, we need to determine the ratio in which FeO and Fe203 are formed in the reaction. From the given information, we know that the reaction produces a mixture of FeO and Fe203.
The balanced chemical equation for the reaction is:
4Fe + 3O2 -> 2Fe2O3
The stoichiometric ratio between iron (Fe) and oxygen (O2) is 4:3. Therefore, for every 4 moles of iron consumed, 3 moles of oxygen react.
3. Calculate the moles of oxygen consumed in the reaction:
- The molar mass of oxygen (O2) is 32.00 g/mol.
- The moles of oxygen consumed can be calculated by dividing the mass by the molar mass:
Moles of oxygen consumed = 7.64g / 32.00 g/mol = 0.23875 mol
4. Determine the moles of iron consumed using the stoichiometric ratio:
- Since the stoichiometric ratio between Fe and O2 is 4:3, we can calculate the moles of iron consumed by multiplying the moles of oxygen consumed by the ratio:
Moles of iron consumed = 0.23875 mol x (4 mol Fe / 3 mol O2) = 0.31833 mol
5. Calculate the moles of FeO formed:
- From the balanced equation, we know that for every 4 moles of iron consumed, 2 moles of Fe2O3 are formed.
- Therefore, the moles of FeO formed would be half of the moles of iron consumed:
Moles of FeO = 0.31833 mol / 2 = 0.15916 mol
6. Calculate the moles of Fe203 formed:
- The moles of Fe203 formed would be the remaining moles of iron consumed:
Moles of Fe203 = 0.31833 mol - 0.15916 mol = 0.15916 mol
7. Calculate the masses of FeO and Fe203 formed:
- The molar mass of FeO is 71.85 g/mol.
- The molar mass of Fe203 is 159.69 g/mol.
- The masses of FeO and Fe203 can be calculated by multiplying the respective moles by their molar masses:
Mass of FeO = 0.15916 mol x 71.85 g/mol = 11.44 g
Mass of Fe203 = 0.15916 mol x 159.69 g/mol = 25.46 g
Therefore, in this experiment, approximately 11.44g of FeO and 25.46g of Fe203 were formed.