Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
Hot air rising and cooler air falling.
Answer:
The oxidation number of the metal decreases
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO
Explanation:
When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal
An example of a metal being reduced is;
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.
