# Question #9937f

Dec 21, 2014

The answer is $998$ tonnes of $F {e}_{2} {O}_{3}$ and $225$ tonnes of $C$.

Since this is a multi-step reaction, we'll use the step

$2 {C}_{\left(s\right)} + {O}_{2 \left(g\right)} \to 2 C {O}_{\left(g\right)}$

and the main chemical reaction

$F {e}_{2} {O}_{3 \left(s\right)} + 3 C {O}_{\left(g\right)} \to 2 F {e}_{\left(l\right)} + 3 C {O}_{2 \left(g\right)}$

We know the mass of $F e$ produced, which is given to be $700.0 \cdot {10}^{6} g$ (I've converted it to grams because it will be easier to work with), and $F e$'s molar mass - $55.8 \frac{g}{m o l}$ - which means we can calculate the number of moles

${n}_{F e} = \frac{m}{m o l a r m a s s} = \frac{700.0 \cdot {10}^{6} m o l e s}{55.8 \frac{g}{m o l}} = 1.25 \cdot {10}^{7}$ moles

Since we've got a $1 : 2$ mole ratio between $F {e}_{2} {O}_{3}$ and $F e$, the number of moles of iron (III) oxide will be

${n}_{F {e}_{2} {O}_{3}} = {n}_{F e} / 2 = 6.25 \cdot {10}^{6}$ moles

Iron (III) oxide's molar mass is $59.6 \frac{g}{m o l}$, which means the mass of $F {e}_{2} {O}_{3}$ needed is

${m}_{F {e}_{2} {O}_{3}} = n \cdot m o l a r m a s s = 6.25 \cdot {10}^{6} \cdot 159.6 = 998 \cdot {10}^{6} g$

Since we were dealing with tonnes, this is equal to $998$ tonnes.

In order to determine the amount of coke (C) needed, we need to know how many moles of $C O$ are required for the main reaction; since we have a $2 : 3$ mole-to-mole ratio between $F e$ and $C O$,

$1.25 \cdot {10}^{7} m o l e s F e \cdot \frac{3 m o l e s C O}{2 m o l e s F e} = 1.88 \cdot {10}^{7}$moles

Notice that we have a $1 : 1$ mole-to-mole ratio for $C$ and $C O$ in the first reaction, so

${n}_{C} = {n}_{C O} = 1.88 \cdot {10}^{7}$ moles

Therefore, the mass of $C$ needed is

${m}_{C} = n \cdot m o l a r m a s s = 1.88 \cdot {10}^{7} \cdot 12.0 = 225 \cdot {10}^{6} g$

Again, converting to tonnes will get us $225$ tonnes.

Here's a video detailing the process:

Dec 21, 2014

You will need 1001 t of iron(III) oxide and 226 t of coke.

The balanced equations are

Fe₂O₃ + 3CO → 2Fe + 3CO₂
2C + O₂ → 2CO₂

Iron(III) Oxide

You have to convert mass of Fe → moles of Fe → moles of Fe₂O₃ → mass of Fe₂O₃

We can do this in one long step.

$\text{700 t Fe" × "1000 kg Fe"/"1 t Fe" × "1 kmol Fe"/"55.845 kg Fe" × ("1 kmol Fe"_2"O"_3)/"2 kmol Fe" × ("159.69 kg Fe"_2"O"_3)/("1 mol Fe"_2"O"_3) × ("1 t Fe"_2"O"_3)/("1000 kg Fe"_2"O"_3) = "1001 t Fe"_2"O} 3$

Coke

Here, you have to convert mass of Fe → moles of Fe → moles of CO → moles of C → mass of C

$\text{700 t Fe" × "1000 kg Fe"/"1 t Fe" × "1 kmol Fe"/"55.845 kg Fe" × ("3 kmol CO"_2)/"2 kmol Fe" × "2 kmol C"/("2 kmol CO"_2) × "12.01 kg C"/"1 kmol C" × "1 t C"/"1000 kg C" "= 226 t C}$