# Question 02aab

Apr 15, 2015

You need $\text{1.27 kg}$ of iron (III) oxide to produce that much iron.

Start with the balanced chemical equation for the carbothermic reduction of iron (III) oxide

$F {e}_{2} {O}_{3} + 3 C O \to \textcolor{red}{2} F e + 3 C {O}_{2}$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between iron (III) oxide and iron, which means that, for every mole of the former that reacts, 2 moles of the latter will be produced.

Since you know how many grams of iron were produced by the reaction, you can calculate the number of moles by using iron's molar mass, then work backwards to see how many moles of iron (III) oxide reacted

885cancel("g") * "1 mole Fe"/(55.845cancel("g")) = "15.847 moles Fe"

This means that the number of moles of $F {e}_{2} {O}_{3}$ that reacted is

15.847cancel("moles Fe") * ("1 mole "Fe_2O_3)/(color(red)(2)cancel("moles Fe")) = "7.924 moles " $F {e}_{2} {O}_{3}$

Now use iron (III) oxide's molar mass to see how many grams you have

7.924cancel("moles "Fe_2O_3) * "159.69 g"/(1cancel("mole "Fe_2O_3)) = "1265.3 g "# $F {e}_{2} {O}_{3}$

Expressed in kilograms and rounded to three sig figs, the number of sig figs given for 885 g, the answer will be

$1265.3 \cancel{\text{g") * "1 kg"/(1000cancel("g")) = color(green)("1.27 kg}}$