Question

Iron (III) oxide reacts with carbon to produce solid iron and carbon monoxide. If 690.67 grams of pure iron (III) oxide are used how many grams of iron can be produced?

Answer 1

483.07 grams

Explanations:

The reaction between Iron (III) oxide and carbon to produce solid iron and carbon monoxide is given as

`Fe_2O_3+3C\rightarrow2Fe+3CO`

Determine the moles of Iron(III) oxide

`\begin{gathered} moles\text{ of Fe}_2O_3=\frac{mass}{molar\text{ mass}} \\ moles\text{ of Fe}_2O_3=(690.67)/(159.69) \\ moles\text{ of Fe}_2O_3=4.325moles \end{gathered}`

According to stoichiometry, 1 mole of Iron(III)oxide produced 2 moles of Iron. The moles of Iron required is expressed as:

`\begin{gathered} mole\text{ of Fe}=2*4.325moles \\ mole\text{ of Fe}=8.65moles \end{gathered}`

Determine the mass of Iron produced

`\begin{gathered} Mass\text{ of Fe}=mole* molar\text{ mass} \\ Mass\text{ of Fe}=8.65*55.845 \\ Mass\text{ of Fe}=483.07grams \end{gathered}`

Hence the mass of iron that can be produced is 483.07 grams