Question

2. Molten Iron metal and carbon monoxide are produced in a blast furnace by the reactionof iron(III) oxide and pure carbon. If 70.0 grams of Fe203 are used, how many grams ofiron can be produced?The Balanced reaction is:

Answer 1

The Balanced reaction is: Fe₂O₃ + 3C → 2Fe + 3CO

The grams of iron (Fe) produced = 48.96 grams.

Step-by-step explanation

Given:

Mass of Fe₂O₃ used = 70.0 grams

What to find:

The grams of iron (Fe) produced.

Step-by-step solution:

The balanced reaction between iron(III) oxide and pure carbon in a blast furnace to produce molten iron metal and carbon monoxide is:

Fe₂O₃ + 3C → 2Fe + 3CO

The next step is to convert 70.0 grams of Fe₂O₃ to moles using the mole formula

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ The\text{ }molar\text{ }mass\text{ }of\text{ }Fe₂O₃=159.69\text{ }g\text{/}mol \\ \\ Moles=\frac{70.0\text{ }g}{159.69\text{ }g\text{/}mol} \\ \\ Moles=0.438349301\text{ }mol\text{ }Fe₂O₃ \end{gathered}`

Now, we can use the moles of Fe₂O₃ and the mole ration from the equation above to determine the moles of Fe produced.

`\begin{gathered} 1\text{ }mol\text{ }Fe₂O₃=2\text{ }mol\text{ }Fe \\ \\ 0.438349301\text{ }mol\text{ }Fe₂O₃=x\text{ }mol\text{ }Fe \\ \\ x=\frac{0.438349301\text{ }mol\text{ }Fe₂O₃}{1\text{ }mol\text{ }Fe₂O₃}*2\text{ }mol\text{ }Fe \\ \\ x=0.876698603\text{ }mol\text{ }Fe \end{gathered}`

Finally, we can convert 0.876698603 mol Fe to mass in grams as follows

`\begin{gathered} Mass=Moles* Molar\text{ }mass \\ \\ Molar\text{ }mass\text{ }of\text{ }Fe=55.845\text{ }g\text{/}mol \\ \\ Mass=0.876698603\text{ }mol*55.845\text{ }g\text{/}mol \\ \\ Mass=48.96\text{ }g \end{gathered}`

The grams of iron (Fe) produced is 48.96 grams.