Question

Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO.

Fe2O3 + 3CO --> 2Fe + 3CO2

In the process, 213 g of Fe2O3 are reacted with 140 of CO.
(a) calculate mass of Fe formed
(b) how many moles of the excess reagent is left at the end of the reaction?

Pls show me how to calculate this using dimensional analysis :(

Answer 1

ok so you already have balanced chemical equation... so next step would be to write down what you have and what you need:

Fe2O3 + 3CO --> 2Fe + 3CO2

m(Fe2O3)=213 g
m(CO)=140 g
---------------
n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?

so now all you need to do is to get to it...

n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol

n(CO)= m(CO) / M(CO)
n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol

now lets see what's the ratio in which reactants react to give products...
(I have to go out for cca 15 min and when i get back i will finish this...)

Answer 2

i dont understand.. :(

Answer 3

im back, can you tell me which part you dont understand so i can clear it up...

Answer 4

how do i calculate the limiting reactant? im very much confuse with everything now.

Answer 5

we are coming to that in next step...

Answer 6

okay, go on.

Answer 7

ok so now you need to get limiting reagent... to me it is obvious but that comes from experience, now how do you confirm which one is limiting reactant? you can calculate for any product what amount would you get from all reactants with chemical reaction ratios in mind... this sounds a bit confusing but let me explain by calculation and i hope you will get it...

from reaction you get ratios (example 1):

n(Fe) / n(Fe2O3) = 2/1 = 2
n(Fe) = 2 * n(Fe2O3)
n(Fe) = 2 * 1,33 mol = 2,66 mol

and

n(Fe) / n(CO) = 2/3
n(Fe) = 2/3 * n(CO)
n(Fe) = 2/3 * 4,99 mol = 3,32 mol

now you can see that limiting reagent is Fe2O3 cause you get less product!

now lets do second example (example 2):

n(CO2) / n(Fe2O3) = 3/1 = 3
n(CO2) = 3 * n(Fe2O3)
n(CO2) = 3 * 1,33 mol = 3,99 mol

and

n(CO2) / n(CO) = 3 / 3 = 1
n(CO2) = n(CO)
n(CO2) = 4,99 mol

and here too you get that Fe2O3 is limiting reagent cause less product is produced!
do you get it?

Answer 8

i get it! so to calculate the limiting reactant i just go directly from the mass of the reactant to the mol of the product? is that correct?

Answer 9

yes and to calculate limiting reagent you ALWAYS and i mean ALWAYS calculate with MOLES!!! not with mass or concentration but with moles! :D

Answer 10

do you know how to continue?

Answer 11

okay thnks! and how do i calculate the excess?

Answer 12

now that you know that Fe2O3 is limiting reagent you know that excess will be CO and you just calculate with Fe2O3 how much product will there be and reverse the process, calculate from that amount that you got from Fe2O3 how much CO is needed and substract from 4,99 moles...
get it?

Answer 13

no...

Answer 14

you mean calculate the mole of CO using mole of Fe2O3?

Answer 15

then minus it with the originally mole of CO obtain?

Answer 16

yes and no, calculate moles of CO2 and Fe from Fe2O3 since Fe2O3 is limiting reagent and from that moles of for example CO2 you calculate how much CO is needed. you already have moles of CO2 since we calculated it for limiting reagent!

n(CO2) = 3,99 mol

now you calculate reversing sides like this:

n(CO) / n(CO2) = 3/3 = 1
n(CO) = n(CO2)
n(CO) = 3,99 mol

and excess is:

n(CO, excess) = 4,99 mol - 3,99 mol = 1 mol

get it?

Answer 17

yes i think i get it...

Answer 18

thank you!

Answer 19

not a problem, glad to help... ;)