how many moles of Fe2O2 would be formed by the action of Oxygen one kilogram of iron?
Here is a link that will give you a worked example of a stoichiometry problem. Just follow the steps. I suggest you look at your problem and the post; I suspect you want Fe2O3 and not Fe2O2.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of moles of Fe2O3 formed by the action of oxygen on one kilogram of iron, we need to consider the balanced chemical equation for the reaction:
4Fe + 3O2 -> 2Fe2O3
From the balanced equation, we can see that 4 moles of iron (Fe) react with 3 moles of oxygen (O2) to produce 2 moles of iron(III) oxide (Fe2O3).
To calculate the number of moles of iron, we need to know the molar mass of iron (Fe), which is approximately 55.845 g/mol.
First, let's convert the mass of iron (1 kilogram) to grams:
1 kilogram = 1000 grams
Next, we can calculate the number of moles of iron using its molar mass:
Number of moles of iron = mass of iron (in grams) / molar mass of iron
Number of moles of iron = 1000 g / 55.845 g/mol
Number of moles of iron = 17.88 mol (approx.)
According to the balanced chemical equation, 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron(III) oxide. Therefore, the number of moles of Fe2O3 formed would be:
Number of moles of Fe2O3 = (Number of moles of iron / 4) * 2
Number of moles of Fe2O3 = (17.88 mol / 4) * 2
Number of moles of Fe2O3 = 8.94 mol (approx.)
Hence, the action of oxygen on one kilogram of iron would form approximately 8.94 moles of Fe2O3.