Question

If 12.52mol of aluminum metal is reacted with2.38mol of iron (III) oxide, how many particles, inmoles of iron metal are likely to form?Reaction: 2Al(s) + Fe₂O3(aq)--> Al2O3(aq) + 2Fe(s)

Answer 1

4.76 mol Fe

Procedure

Consider the following balanced equation and determine the limiting reagent using the coefficients method

2Al(s) + Fe₂O3(aq)--> Al2O3(aq) + 2Fe(s)

`12.52\text{ mol Al}\frac{2\text{ mol Fe}}{2\text{ mol Al}}=12.52\text{ mol Fe}`
`2.38\text{ mol Fe}_2\text{O}_3\frac{2\text{ mol Fe}}{1\text{ mol Fe}_2\text{O}_3}=4.76\text{ mol Fe}`

The lowest amount is the one produced by the iron (III) oxide therefore that is the limiting reagent and 4.76 mol of Fe is the max amount that can be produced of Iron.