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Because C and Fe appear in uncombined, atomic form in this equation we can ignore them initially and just look at the Fe2O3 and CO2. There are an odd number of atoms of O in the Fe2O3 and a odd number in the CO2. If I just double the number of molecules of Fe2O3 going into the reaction then this will also double the number of atoms of oxygen, thereby making the number even(which is what I want, eh). Here's what I have so far:

2 Fe2O3 + C -> Fe + CO2

Because there are six oxygen atoms on the left there must be six on the right. Therefore there are three CO2 molecules:

2 Fe2O3 + C -> Fe + 3 CO2

Finally:

  • Four Fe's going in means four must come out.
  • Three C's coming out means three must go in.

2 Fe2O3 + 3 C -> 4 Fe + 3 CO2

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βˆ™ 11y ago
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βˆ™ 6mo ago

To balance the equation Fe2O3 + 3C β†’ 2Fe + 3CO2, you can first balance the carbon atoms by putting a coefficient of 3 in front of CO2. Next, balance the oxygen atoms by adding a coefficient of 4 in front of Fe2O3. The balanced equation is 4Fe2O3 + 3C β†’ 8Fe + 9CO2.

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Q: How do you balance the equation Fe2O3 plus C changed to FetCO2?
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