Question

Question #9937f

Answer 1

The answer is `998` tonnes of `Fe_2O_3` and `225` tonnes of `C`.

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

`2C_((s)) + O_(2(g)) -> 2CO_((g))`

and the main chemical reaction

`Fe_2O_(3(s)) + 3CO_((g)) -> 2Fe_((l)) + 3CO_(2(g))`

We know the mass of `Fe` produced, which is given to be `700.0 * 10^6g` (I've converted it to grams because it will be easier to work with), and `Fe`'s molar mass - `55.8g/(mol)` - which means we can calculate the number of moles

`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` mole ratio between `Fe_2O_3` and `Fe`, the number of moles of iron (III) oxide will be

`n_(Fe_2O_3) = n_(Fe)/2 = 6.25 * 10^6` moles

Iron (III) oxide's molar mass is `59.6 g/(mol)`, which means the mass of `Fe_2O_3` needed is

`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` tonnes.

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

`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` mole-to-mole ratio for `C` and `CO` in the first reaction, so

`n_(C) =n_(CO) = 1.88 * 10^7` moles

Therefore, the mass of `C` needed is

`m_(C) = n * molarmass = 1.88 * 10^7 * 12.0 = 225 * 10^6g`

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

Here's a video detailing the process:

Answer 2

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" × "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" × "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"`