iron (III) oxide is formed when iron combines with oxygen in the air. how many grams of Fe2O3 are formed when 16.7g of Fe reacts completely with oxygen? 4Fe(s) + 3O2(grams) => 2Fe2O3(s)
23.9 g
All of these stoichiometry problem are about the same. Remember these four steps.
Step 1. Write the balanced equation. You have that.
4Fe + 3O2 ==> 2Fe2O3
Step 2. Convert what you have (in this case g Fe) to mols remembering that mols = g/molar mass.
mols Fe = 16.7 g/55.8 = 0.299 mols Fe.
Step 3. Convert mols of what you have (in this case mols Fe) to mols of what you want (in this case mols Fe2O3) using the coefficients in the balanced equation from step 1.
mols Fe2O3 = mols Fe x (2 mols Fe2O3/4 mols Fe) = 0.299 x (2/4) = 0.299 x 1/2 = 0.150
Step 4. Now convert mols of Fe2O3 to grams usiong the equation from step 2 (but rearranged). There we had mols = g/molar mass. Rearrange to grams = mols x molar mass = 0.150 x molar mass Fe2O3 = xx g Fe2O3.
Check my work.
Post your work if you get stuck.
Well, well, well! It seems like our friend iron (Fe) is getting into a heartwarming relationship with the charming oxygen (O2) from the air. But, before we dive into calculations, let me spin some clownish wisdom for you:
Why did the iron go to therapy? Because it felt a bit rusty in its relationships and needed some oxygen to spice things up!
Now, let's get down to business and calculate the number of grams of Fe2O3 formed. First, we need to find the molar mass of Fe2O3:
Molar mass of Fe2O3 = (2 * molar mass of Fe) + (3 * molar mass of O)
= (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
= 111.69 g/mol + 48.00 g/mol
= 159.69 g/mol
Now, we can use the stoichiometry of the reaction to find the grams of Fe2O3 formed when 16.7 g of Fe reacts completely:
First, we need to find the number of moles of Fe:
moles of Fe = mass of Fe / molar mass of Fe
= 16.7 g / 55.845 g/mol
≈ 0.299 mol
According to the balanced equation, the mole ratio between Fe and Fe2O3 is 4:2. So, the number of moles of Fe2O3 formed will be half the number of moles of Fe:
moles of Fe2O3 = moles of Fe / 2
= 0.299 mol / 2
= 0.1495 mol
Finally, we can calculate the mass of Fe2O3 formed using the molar mass of Fe2O3:
mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3
= 0.1495 mol * 159.69 g/mol
≈ 23.87 g
So, approximately 23.87 grams of Fe2O3 are formed when 16.7 grams of Fe reacts completely with oxygen. Now, isn't it amazing how chemistry can bring elements together?
To find the number of grams of Fe2O3 formed, we can use the stoichiometry of the balanced chemical equation:
4Fe(s) + 3O2(g) -> 2Fe2O3(s)
According to the balanced equation, it takes 4 moles of Fe to react with 3 moles of O2 to produce 2 moles of Fe2O3.
First, we need to convert the mass of Fe (16.7g) to moles. The molar mass of Fe is 55.85 g/mol, so:
16.7 g Fe * (1 mol Fe / 55.85 g Fe) = 0.298 mol Fe
Since the stoichiometry ratio of Fe to Fe2O3 is 4:2, we can set up a proportion to find the number of moles of Fe2O3 formed:
0.298 mol Fe / 4 mol Fe2O3 = x mol Fe2O3 / 2
Cross-multiplying the proportion:
0.298 mol Fe * 2 mol Fe2O3 = 4 mol Fe2O3 * x mol Fe2O3
0.596 mol Fe2O3 = 4x mol Fe2O3
Dividing both sides by 4:
0.596 mol Fe2O3 / 4 = x mol Fe2O3
0.149 mol Fe2O3 = x mol Fe2O3
Finally, we need to convert moles of Fe2O3 to grams. The molar mass of Fe2O3 is:
(2 mol Fe * 55.85 g/mol) + (3 mol O * 16.00 g/mol) = 159.69 g/mol
Now, we can calculate the mass of Fe2O3 formed:
0.149 mol Fe2O3 * (159.69 g Fe2O3 / 1 mol Fe2O3) = 23.81 g Fe2O3
Therefore, 16.7 grams of Fe will produce 23.81 grams of Fe2O3.
To determine the number of grams of Fe2O3 formed, we need to use stoichiometry and convert the given mass of Fe to the mass of Fe2O3 using the balanced chemical equation.
First, we need to determine the molar mass of Fe2O3.
Fe has a molar mass of 55.845 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Since Fe2O3 has two iron atoms and three oxygen atoms, the molar mass of Fe2O3 is:
(2 * 55.845) + (3 * 16.00) = 159.69 g/mol.
Next, we can calculate the number of moles of Fe using its molar mass. We can use the given mass of 16.7 g and the molar mass of Fe (55.845 g/mol) to find the number of moles:
Number of moles of Fe = mass / molar mass = 16.7 g / 55.845 g/mol = 0.298 mol.
Now, we can use stoichiometry to determine the number of moles of Fe2O3 formed. According to the balanced equation, 4 moles of Fe react to produce 2 moles of Fe2O3. Therefore, the molar ratio between Fe and Fe2O3 is 4:2, or 2:1.
Number of moles of Fe2O3 = (0.298 mol Fe) * (2 mol Fe2O3 / 4 mol Fe) = 0.149 mol Fe2O3.
Finally, we can convert the number of moles of Fe2O3 to grams using the molar mass of Fe2O3:
Mass of Fe2O3 = number of moles * molar mass = 0.149 mol * 159.69 g/mol = 23.74 g.
Therefore, when 16.7 g of Fe reacts completely with oxygen, 23.74 g of Fe2O3 is formed.