# Given 200*g masses each of sodium metal, and ferric oxide.....can you address..?

## ...which reagent is in excess in the context of iron reduction....? $F {e}_{2} {O}_{3} \left(s\right) + 6 N a \left(s\right) \rightarrow 3 N {a}_{2} O \left(s\right) + 2 F e \left(s\right)$

Mar 25, 2017

You have the stoichiometric equation:..........

#### Explanation:

$6 N a + F {e}_{2} {O}_{3} \rightarrow 3 N {a}_{2} O + 2 F e$.

$\text{Moles of natrium}$ $=$ $\frac{200 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 8.70 \cdot m o l$

$\text{Moles of ferric oxide}$ $=$ $\frac{200 \cdot g}{159 , 69 \cdot g \cdot m o {l}^{-} 1} = 1.25 \cdot m o l$

Given the $6 : 1$ stoichiometry, the natrium is in excess, and all the ferric oxide should be reduced to the metal.

Note that this would be an extremely expensive and impractical way of producing iron metal. Far better (and miles cheaper) to produce iron the traditional way, and heat it up with coke (i.e. carbon), and reduce the iron while producing $C O$ and $C {O}_{2}$:

$F {e}_{2} {O}_{3} + 3 C \rightarrow 2 F e + 3 C O \uparrow$

Are these redox equations?