Question #9937f

2 Answers
Dec 21, 2014

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:

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.

#"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"#