Question #9937f

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Question #9937f

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Answer 1 · 2014-12-21T16:30:02

The answer is $998$ $998$ tonnes of $F e_{2} O_{3}$ $Fe_2O_3$ and $225$ $225$ tonnes of $C$ $C$ .

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

$2 C_{(s)} + O_{2 (g)} \to 2 C O_{(g)}$ $2C_((s)) + O_(2(g)) -> 2CO_((g))$

and the main chemical reaction

$F e_{2} O_{3 (s)} + 3 C O_{(g)} \to 2 F e_{(l)} + 3 C O_{2 (g)}$ $Fe_2O_(3(s)) + 3CO_((g)) -> 2Fe_((l)) + 3CO_(2(g))$

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

$n_{F e} = \frac{m}{m o l a r m a s s} = \frac{700.0 \cdot 10^{6} m o l e s}{55.8 \frac{g}{m o l}} = 1.25 \cdot 10^{7}$ $n_(Fe) = m/(molar mass) = (700.0 * 10^6 mol es)/(55.8 g/(mol)) = 1.25 * 10^7$ moles

Since we've got a $1 : 2$ $1:2$ mole ratio between $F e_{2} O_{3}$ $Fe_2O_3$ and $F e$ $Fe$ , the number of moles of iron (III) oxide will be

$n_{F e_{2} O_{3}} = \frac{n_{F e}}{2} = 6.25 \cdot 10^{6}$ $n_(Fe_2O_3) = n_(Fe)/2 = 6.25 * 10^6$ moles

Iron (III) oxide's molar mass is $59.6 \frac{g}{m o l}$ $59.6 g/(mol)$ , which means the mass of $F e_{2} O_{3}$ $Fe_2O_3$ needed is

$m_{F e_{2} O_{3}} = n \cdot m o l a r m a s s = 6.25 \cdot 10^{6} \cdot 159.6 = 998 \cdot 10^{6} g$ $m_(Fe_2O_3) = n * molarmass = 6.25 * 10^6 * 159.6 = 998 * 10^6g$

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

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

$1.25 \cdot 10^{7} m o l e s F e \cdot \frac{3 m o l e s C O}{2 m o l e s F e} = 1.88 \cdot 10^{7}$ $1.25 * 10^7 mol es Fe * (3 mol es CO)/(2 mol es Fe) = 1.88 * 10^7$ moles

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

$n_{C} = n_{C O} = 1.88 \cdot 10^{7}$ $n_(C) =n_(CO) = 1.88 * 10^7$ moles

Therefore, the mass of $C$ $C$ needed is

$m_{C} = n \cdot m o l a r m a s s = 1.88 \cdot 10^{7} \cdot 12.0 = 225 \cdot 10^{6} g$ $m_(C) = n * molarmass = 1.88 * 10^7 * 12.0 = 225 * 10^6g$

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

Here's a video detailing the process:

Answer 2 · 2014-12-21T16:56:38

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.

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

Coke

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

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

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