What mass of O2 is needed to react with 10.2 grams of Fe?
The reaction of iron with oxygen can produce either `FeO` or `Fe_2O_3` , depending on the temperature and other conditions. I will show you the calculation for both.
This is a stoichiometry problem. It can be solved by using dimensional analysis (conversion factors) to convert from grams of Fe to moles of Fe to moles of O2 to grams of O2. The conversion factors you need are the molar masses of Fe and O2 and the mole ratio of O2 to Fe, which is the coefficients from the balanced equation. The mole ratio is different for the two different equations.
Here's the equation for the more common of the two, the production of Fe2O3, which occurs when iron rusts:
`4 Fe + 3 O_2 -> 2 Fe_2O_3`
Start with the given quantity, 10.2g Fe:
10.2g Fe x (1 mol Fe/55.8g Fe) x (3 mol O2/4 mol Fe) x 32.0g O2/1 mol O2
= 4.39 grams O2 needed
Now for the FeO:
`2 Fe + O2 -> 2 FeO`
The only difference is the mole ratio of O2 to Fe:
10.2g Fe x (1mol Fe/55.8g Fe) x (1 mol O2/2 mol Fe) x (32.0g O2/1 mol O2)
= 2.92 grams O2 needed
Be sure to include units when using dimensional analysis to solve problems. If you end up with the wrong units for your answer you'll know to go back and check your work.