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

Question

1370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-12T12:46:31

Iron is the reagent in slight excess.

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#.

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?