Question

How many moles of rust `(Fe_2O_3)` will be formed by the rusting of 25.0 grams of iron (`Fe`)?

Answer 1

Approx. `1/4` `mol` of ferric oxide.

Explanation:

`2Fe(s) + 3/2O_2(g) rarr Fe_2O_3(aq)`

Moles of iron `=` `(25.0*g)/(55.85*g*mol^-1)` `=` `??` `"mol"`

From the stoichiometry of the reaction, HALF this molar quantity of rust will be formed. What is the mass of rust?