Question

A `6*g` mass of iron reacts with a `1.6*g` mass of dioxygen to form `FeO`. What is the reagent in excess, and what is the mass of product formed?

Answer 1

Iron is the reagent in slight excess.

Explanation:

We need a stoichiometric equation:

`Fe(s) + 1/2O_2(g) stackrel(Delta)rarr FeO(s)`

`"Moles of iron"=(6*g)/(55.8*g*mol^-1)=0.108*mol`

`"Moles of dioxygen"=(1.6*g)/(32.0*g*mol^-1)=0.050*mol`

Clearly, the metal is present in slight stoichiometric excess, i.e. `8xx10^-3*mol`.