Hmk. Problem H.11 (Fundamentals)

Moderators: Chem_Mod, Chem_Admin

Samudrala_Vaishnavi 3A
Posts: 98
Joined: Wed Sep 30, 2020 9:34 pm

Hmk. Problem H.11 (Fundamentals)

Postby Samudrala_Vaishnavi 3A » Mon Oct 05, 2020 4:44 pm

Question: In one stage in the commercial production of iron metal in a blast furnace, the iron(III) oxide, Fe 2 O 3 , reacts with carbon monoxide to form solid Fe 3 O 4 and carbon dioxide gas. In a second stage, the Fe 3 O 4 reacts further with carbon monoxide to produce solid elemental iron and carbon dioxide. Write the balanced equation for each stage in the process. (the numbers in the side are subscripts btw).

I was able to create the equations and balance the first equation, but I'm stuck specifically on the problem when looking/ balancing at the second equation once I balanced the oxygens:

I got: (when I balanced the oxygens)

Whenever I continue after this step it gets confusing. I keep having to rebalance the oxygens or carbons. Is there a way to rebalance the carbons without messing up everything or a way to finish balancing this without having a mess?

Nathan Chu 3H
Posts: 96
Joined: Wed Sep 30, 2020 9:48 pm

Re: Hmk. Problem H.11 (Fundamentals)

Postby Nathan Chu 3H » Mon Oct 05, 2020 4:52 pm

Instead of using 2 moles Fe3O4, try using one mol to balance the equation. It should yield a balanced equation of Fe3O4(s) + 4 CO(g) -> 3 Fe(s) + 4 CO2(g).

Janelle Gokim 3B
Posts: 115
Joined: Wed Sep 30, 2020 9:42 pm
Been upvoted: 3 times

Re: Hmk. Problem H.11 (Fundamentals)

Postby Janelle Gokim 3B » Mon Oct 05, 2020 4:52 pm

Hi! I understand your struggle with this, I tend to go the guess and check method too when balancing and often have to resort to changing my numbers. I just balanced it right now and got Fe3O4 + 4CO --> 3Fe +4CO2. I wish I could be more help with a different method but what I usually do is avoid having odd numbers of certain elements. I saw that you put 5 as a coefficient so this in turn would make it impossible to balance the C on the other side without creating odd values for other elements. I think just being conscious of this and experimenting with different numbers is the best bet.
Last edited by Janelle Gokim 3B on Mon Oct 05, 2020 4:55 pm, edited 1 time in total.

Melody Wu 2L
Posts: 109
Joined: Wed Sep 30, 2020 10:00 pm
Been upvoted: 1 time

Re: Hmk. Problem H.11 (Fundamentals)

Postby Melody Wu 2L » Mon Oct 05, 2020 4:53 pm

The answer is actually:


I kind of had to play around with the numbers to get it right. I balanced Fe first and then started with adding a stoichiometric coefficient of 2 on CO2 and CO. Then I did a bit of trial and error and increased it to 3, then 4, until it worked. For ones like these I think you just have to mess around a few times.

Jeffrey Doeve 2I
Posts: 103
Joined: Wed Sep 30, 2020 9:51 pm

Re: Hmk. Problem H.11 (Fundamentals)

Postby Jeffrey Doeve 2I » Mon Oct 05, 2020 4:55 pm

So the way I approached this problem is based on first getting the big picture of O atoms. To balance the O atoms, you have to first realize that the number of O atoms on the product side can only be even(CO2), where as the number of O atoms on the reactant side can be either odd or even(Fe3O4 and CO). So now, we know that the stoichiometric coefficient has to be an even number for CO. Starting to actually balance the equation, we start by looking at Fe in Fe3O4. Since there are 3 Fe atoms here, we put the stoichiometric coefficient of 3 in front of the Fe on the product side. Since CO results in an odd number of O atoms, we know the coefficient must be even and the same as the coefficient of CO2. Thus, we put the coefficient of 4 in front of both CO and CO2. This fixes are O balancing dilemma with 8 O atoms on each side.
Final answer: Fe3O4(s) + 4CO -> 3Fe(s) + 4CO2(g)

Samudrala_Vaishnavi 3A
Posts: 98
Joined: Wed Sep 30, 2020 9:34 pm

Re: Hmk. Problem H.11 (Fundamentals)

Postby Samudrala_Vaishnavi 3A » Mon Oct 05, 2020 5:33 pm

Jeffrey Doeve 3H wrote:So the way I approached this problem is based on first getting the big picture of O atoms. To balance the O atoms, you have to first realize that the number of O atoms on the product side can only be even(CO2), where as the number of O atoms on the reactant side can be either odd or even(Fe3O4 and CO). So now, we know that the stoichiometric coefficient has to be an even number for CO. Starting to actually balance the equation, we start by looking at Fe in Fe3O4. Since there are 3 Fe atoms here, we put the stoichiometric coefficient of 3 in front of the Fe on the product side. Since CO results in an odd number of O atoms, we know the coefficient must be even and the same as the coefficient of CO2. Thus, we put the coefficient of 4 in front of both CO and CO2. This fixes are O balancing dilemma with 8 O atoms on each side.
Final answer: Fe3O4(s) + 4CO -> 3Fe(s) + 4CO2(g)


This makes more sense, thank you for the detailed explanation. I guess I have to avoid odd numbered coefficients when dealing with odd numbers of atoms as best as possible and look at the big picture the most.


Return to “Balancing Chemical Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest