O does not exist by itself. The balanced equation would be 4Fe + 3O2 --> 2Fe2O3
The oxidation number of Fe in FeO (iron oxide) is +2. Oxygen is typically assigned an oxidation number of -2, so since FeO is a neutral compound, the oxidation number of Fe must be +2 to balance out the charge of the oxygen.
FeO is 50 mole % Fe and 50 mole % O FeO is 77.73 mass % Fe and 22.27 mass % O mass fraction X = molar mass X / (total molar mass of compound) mass % Fe = [atomic mass Fe] / ([atomic mass Fe] + [atomic mass O])
Because its a compound that only forms when Iron (Fe) bonds with oxygen (O) to form FeO.
First of all 'FeO' is NOT an element. It is a compound composed of two elements. The elements being iron(Fe) and oxygen(O). Next, you appear to misunderstand between 'Ionic' and 'molecular'. Any substance is a composed of two or more atoms is a molecule. The method of combination of these atoms can be either 'Ionic' or 'covalent'. Iron(II) Oxide (FeO) is an ionic molecule (compound). Fe^(2+)(aq) + O^(2-)(aq) = FeO(s)
Yes, FeO is connected by an ionic bond. In FeO, iron (Fe) has a positive charge and oxygen (O) has a negative charge, leading to the formation of an ionic bond between the two elements.
The formula FeO represents one iron (Fe) atom and one oxygen (O) atom, totaling two atoms in the compound.
FeO is an ionic bond. Iron (Fe) is a metal and oxygen (O) is a non-metal, so they tend to form an ionic bond where Fe loses electrons to form Fe2+ cation and O gains electrons to form O2- anion.
To calculate the amount of iron that can be recovered from FeO, we need to consider the molar ratio between Fe and FeO. FeO consists of 1 iron atom and 1 oxygen atom. The molar mass of Fe is 55.85 g/mol, and the molar mass of FeO is 71.85 g/mol. By dividing the molar mass of Fe by the molar mass of FeO and multiplying by the given mass of FeO, we can determine the amount of iron that can be recovered.
The reactions are:4 Fe + 3 O2 + 2 H2O = 4 FeO(OH)2 FeO(OH) -------Fe2O3 + H2O
In the balanced reaction ( \text{Fe}_2\text{S}_3 + 4\text{O}_2 \rightarrow 2\text{FeO} + 3\text{SO}_2 ), the mole ratio of ( \text{Fe}_2\text{S}_3 ) to ( \text{O}_2 ) is 1:4. This means that for every 1 mole of iron(III) sulfide (( \text{Fe}_2\text{S}_3 )), 4 moles of oxygen (( \text{O}_2 )) are required for the reaction to proceed.
feo2 feo3 feo4
FeO