To answer this question we will have to know which of the many oxides of iron you are targeting.
in one molecule of Fe2O3 there would be 3 oxygen atoms
The molar mass of anhydrous iron(II) nitrate is 179,91 g.
The mass percent of oxygen is approximately 21% in Earth's atmosphere.
The change in mass is 0.225 g. The oxygen reacted can be calculated by determining the difference in the amount of oxygen in the reactants and products. The molar ratio of Fe2O3 to O2 is 2:3, so the mass of oxygen reacted would be 0.150 g.
Adding together the mass of two irons and three oxygen.....,251 grams Fe2O3 (1 mole Fe2O3/159.7 grams)= 1.57 moles iron II oxide ( also known as ferric oxide )===================================
in one molecule of Fe2O3 there would be 3 oxygen atoms
There are two atoms of iron and three atoms of oxygen in the compound Fe2O3. The mass percent of iron is 69.943%
The molar mass of anhydrous iron(II) nitrate is 179,91 g.
The mass percent of oxygen is approximately 21% in Earth's atmosphere.
The change in mass is 0.225 g. The oxygen reacted can be calculated by determining the difference in the amount of oxygen in the reactants and products. The molar ratio of Fe2O3 to O2 is 2:3, so the mass of oxygen reacted would be 0.150 g.
The percent by mass of oxygen in N2O4 is 69,56 %.
Adding together the mass of two irons and three oxygen.....,251 grams Fe2O3 (1 mole Fe2O3/159.7 grams)= 1.57 moles iron II oxide ( also known as ferric oxide )===================================
The oxide Fe2O3 has 3 oxygen atoms in the molecule.
You need 150,41 g oxygen.
To find the mass percent of oxygen in the compound, first calculate the total mass of the compound by adding the masses of magnesium and oxygen: 14.5 g + 3.5 g = 18.0 g. Then, use the formula for mass percent: (mass of oxygen / total mass) × 100%. Thus, the mass percent of oxygen is (3.5 g / 18.0 g) × 100% ≈ 19.44%.
To determine the mass of Fe2O3 formed, you need to first find the molar mass of Fe and Fe2O3. Then, use stoichiometry to calculate the mass of Fe2O3 that can be produced from 16.7 g of Fe. The balanced equation is 4Fe + 3O2 → 2Fe2O3, which shows that 4 mol of Fe produces 2 mol of Fe2O3. By converting 16.7 g of Fe to moles, you can then use the mole ratio to find the grams of Fe2O3 formed.
Ferric oxide often refers to iron (III) oxide, which has a chemical formula Fe2O3. The atomic mass of iron (Fe) is 55.85 g/mol, and the atomic mass of oxygen (O) is 16.00 g/mol. Therefore, the molar mass of Fe2O3 is 159.69 g/mol. To calculate the percentage composition of iron (III) oxide, divide the atomic mass of iron by the molar mass of the compound and multiply by 100. So, (2 * 55.85) / 159.69 ≈ 0.698 or 69.8% is the percent of ferric oxide.