# A student conducts the following reaction: Fe_3O_4 + 4 CO \rightarrow 3 Fe + 4CO_2. If the...

## Question:

A student conducts the following reaction:

{eq}Fe_3O_4 + 4 CO \rightarrow 3 Fe + 4CO_2 {/eq}.

If the student starts with 145 g {eq}Fe_3O_4 {/eq}, what is the percent yield of the reaction if only 38 liters of {eq}CO_2 {/eq} were produced?

## Iron (II,III) Oxide:

The molecule {eq}Fe_3O_4 {/eq} is called iron (II,III) oxide because one of the iron atoms has a +2 oxidation state and two of the iron atoms have a +3 oxidation state. It is sometimes written with the formula {eq}FeO \cdot Fe_2O_3 {/eq} and called ferrous ferric oxide.

## Answer and Explanation: 1

According to the balanced chemical reaction equation, every mole of iron (II,III) oxide reacts in excess carbon monoxide to produce 4 moles of carbon dioxide. The mass of iron (II,III) oxide is converted to a quantity of moles by dividing by its molar mass (231.53 g/mol). The resulting quantity of moles of carbon dioxide can be converted to a volume by multiplying by the molar mass of carbon dioxide (44.01 g/mol) and dividing by its density (1.977 g/L)

{eq}145\ g\ \mathrm{Fe_3O_4} \left(\dfrac{1\ mol\ \mathrm{Fe_3O_4}}{231.53\ g\ \mathrm{Fe_3O_4}}\right) \left(\dfrac{4\ mol\ \mathrm{CO_2}}{1\ mol\ \mathrm{Fe_3O_4}}\right) \left(\dfrac{44.01\ g\ \mathrm{CO_2}}{1\ mol\ \mathrm{CO_2}}\right) \left(\dfrac{1\ L\ \mathrm{CO_2}}{1.977\ g\ \mathrm{CO_2}}\right) = 55.8\ L\ \mathrm{CO_2} {/eq}

This value is called the theoretical yield, and the percent yield of the reaction is the ratio of the actual yield from running the experiment to this theoretical yield found by stoichiometry:

{eq}\begin{align} \text{% yield} &= \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \\ &= \dfrac{38\ L}{55.8\ L} \times 100\% \\ \\ &= \boxed{\ 68\%\ } \end{align}{/eq}