Question

Rusting of iron is a very common chemical reaction. It results in one form from Fe reacting with oxygen gas to produce iron (III) oxide. Your sample of iron is 12.0 moles of iron. So which if these is a true statement? NOTE: All numbers located immediately after elemental symbols below should be considered subscripts. A. 4.5 moles of O2 and produce 3.0 moles of Fe2O3. B. 12.0 moles of O2 and produce 24.0 moles of Fe2O3. C. 9.0 moles of O2 and produce 3.0 moles of Fe2O3. D. 9.0 moles of O2 and produce 6.0 moles of Fe2O3 E. none of the above

Answer 1

Answer: The given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

Explanation:

We are given:

Moles of iron = 12.0 moles

The chemical equation for the rusting of iron follows:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

• For oxygen gas:

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 12.0 moles of iron will react with = $\frac{3}{4}\times 12.0=9.0mol$ of oxygen gas

• For iron (III) oxide:

By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 12.0 moles of iron will produce = $\frac{2}{4}\times 12.0=6.0mol$ of iron (III) oxide

Hence, the given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$