Balancing Chemical Reactions  [ENDORSED]

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JTieu_1L
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Balancing Chemical Reactions

Postby JTieu_1L » Thu Oct 01, 2020 1:30 pm

I came across this question while doing the Fundamental problems of H and I do not understand how to balance the second stage chemical reaction:

"In one stage in the commercial production of iron metal in a blast furnace, the iron(III) oxide, Fe2O3, reacts with carbon monoxide to form solid Fe3O4 and carbon dioxide gas. In a second stage, the Fe3O4 reacts further with carbon monoxide to produce solid elemental iron and carbon dioxide. Write the balanced equation for each stage in the process." (#H11, pg F65)

I figured out the equation unbalanced: Fe3O4 + CO = Fe + CO2
However, I can't seem to balance the O and the C. How should I approach this equation?

Lucy_Balish_3G
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Re: Balancing Chemical Reactions

Postby Lucy_Balish_3G » Thu Oct 01, 2020 3:51 pm

You can balance this equation by putting a 4 in front of CO and CO2 and a 3 in front of Fe.
That way you end up with Fe3O4 + 4CO = 3Fe + 4CO2 which is balanced.

SelenaDahabreh1D
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Re: Balancing Chemical Reactions

Postby SelenaDahabreh1D » Thu Oct 01, 2020 4:50 pm

This is how I went about this question:

Skeletal Second Stage Chemical Equation - Fe3O4 + CO --> Fe + CO2

So for balancing the chemical equation for the second stage, I'd start by first balancing Fe. Thus, I'm left with this chemical equation: Fe3O4 + CO--> 3Fe + CO2.
Next, I'd experiment to see how I could get there to be an equal amount of C (Carbon) and O (oxygen) on both sides of the equation. First, I'm going to try to make the same number of Oxygen atoms on both sides.

So on the reactant side (left), I have 5 O, but on the right sight I have 2 O. I would first start by adding a 2 in front of CO on the reactant side. Thus, I will have 6 O on the left. Next, I would add a 3 in front of CO2 on the product side (right). Then, I would be left with 6 O on the left and 6 O on the right.

But then, I'd have an unequal number of C (Carbon) on both sides. Thus, I would try to increase the amount of CO and CO2 on both sides of the reaction.

So, instead of 3CO atoms on the left, I’d experiment with 4CO atoms on the left. Thus, I am left with 8 O atoms on the left and 4 C atoms on the left. To have the same number of C and O atoms on the left side, I’d place a 4 before CO2 on the right side. Then, I'd have 8 O atoms on the left and 8 O atoms on the right, as well as 4 C atoms on the left and 4 C atoms on the right.

Thus, I am left with this balanced chemical equation:
Fe3O4 + 4CO--> 3Fe + 4CO2.

Hope this helps!

Inderpal Singh 2L
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Re: Balancing Chemical Reactions

Postby Inderpal Singh 2L » Sat Oct 03, 2020 3:28 pm

I started out with balancing the Fe, giving me:
Fe3O4 + CO -> 3Fe + CO2.

In an attempt to balance out the oxygens on both sides, I will add a 2 in front of carbon monoxide on the reactant side.
This will give me:
Fe3O4 + 2CO -> 3Fe + CO2.

I will also add a 3 in front of the CO2 on the right in an attempt to balance them out:
Fe3O4 + 2CO -> 3Fe + 3CO2.

We now have equal Fe, 6 oxygen each but Carbon is not balanced with 2 on the reactant side and 3 on the product.
Therefore, I will try to balance them out by adding a 4 to CO2 on the product side and 4 to CO on the reactant side.
Fe3O4 + 4CO -> 3Fe + 4CO2.

Now the equation is balanced : Fe3O4 + 4CO -> 3Fe + 4CO2

My tip is to really take balancing equation one step at a time, it is really easy to mess up, but when you take each element, one step at a time, you will eventually come to a conclusion :)

Sam Wentzel 1F 14B
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Re: Balancing Chemical Reactions

Postby Sam Wentzel 1F 14B » Sat Oct 03, 2020 8:15 pm

_Fe3O4 + _CO -> _Fe + _CO2
I like to approach balancing chemical equations by balancing the most complex molecules first, working towards the least complex last.

I started by adding 1 to Fe3O4, making: Fe3O4 + _CO -> _Fe + _CO2
Next, I balanced Fe by adding a 3 to it, making: Fe3O4 + _CO -> 3Fe + _CO2
Balancing the carbons and oxygens is a bit trickier.

Fe3O4 gives us 4 oxygens, on top of the oxygens added by CO.
Because CO2 on the products side will always have an even number of oxygens regardless of its stoichiometric coefficient, we know that the stoichiometric coefficient of CO must be some even number.

The first even number we can try is 2.
This gives us: Fe3O4 + 2CO -> 3Fe + _CO2
When we try to balance the carbons, we are left with: Fe3O4 + 2CO -> 3Fe + 2CO2
This does not work because we have 6 oxygens on the reactants side, and only 4 on the products side.
Thus the next even number we can try is 4.
This gives us: Fe3O4 + 4CO -> 3Fe + _CO2
When we balance the carbons, we get: Fe3O4 + 4CO -> 3Fe + 4CO2 :D

Chudi Onyedika 3A
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Re: Balancing Chemical Reactions  [ENDORSED]

Postby Chudi Onyedika 3A » Sun Oct 04, 2020 9:03 am

Since you have balanced Fe, move on to another element. There are clearly less O atoms on the product side. Place a different coefficient by the CO2 and analyze how that changes the total number of C and O atoms on each side. If one side has more atoms than the other side, then place a coefficient on the side with less atoms.

Eventually, the reaction balances as: Fe3O4 + 4CO --> 3Fe + 4CO2. This is achieved by placing a coefficient of 4 in front of "CO" and "CO2."

Chem_Mod
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Re: Balancing Chemical Reactions

Postby Chem_Mod » Sun Oct 04, 2020 10:15 am

As I say in class, there are different ways to approach problems that lead to the same correct answer. :-)


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