# 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?

Aug 12, 2017

Iron is the reagent in slight excess.

#### Explanation:

We need a stoichiometric equation:

$F e \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \stackrel{\Delta}{\rightarrow} F e O \left(s\right)$

$\text{Moles of iron} = \frac{6 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1} = 0.108 \cdot m o l$

$\text{Moles of dioxygen} = \frac{1.6 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 0.050 \cdot m o l$

Clearly, the metal is present in slight stoichiometric excess, i.e. $8 \times {10}^{-} 3 \cdot m o l$.