When the following molecular equation is balanced using the smaller possible interger cofficents, the values of these coefficients are;

Question

[ ]CI2(g) + [ ] H2O(I) ——-> [ ] HCI(aq) + [ ] HCIO3(aq)

[ ]Fe(s) + [ ]O2(g) —-> [ ]Fe2O3(s)

the “[ ]” is where I need the answer.

Answer 1

Cl_{2 (g)} + H₂O_(l) → HCl_(aq) + HClO_{3 (aq)}

Start by balancing Oxygen since it is present in only one compound on either side.

Cl_{2 (g)} + 3H₂O_(l) → HCl_(aq) + HClO_{3 (aq)}

Don't forget to multiply the coefficient by the amount in the molecule. Notice how this produces 6 net Hydrogen on the reactant side (3 molecules with 2 H in each)? Since we have already balanced the Oxygen, subtract the amount of Hydrogen in HClO₃ and place the remaining amount of Hydrogen in the HCl

Cl_{2 (g)} + 3H₂O_(l) → 5HCl_(aq) + HClO_{3 (aq)}

Now we have a total of 6 Chlorine on the product side, so adjust the amount on the reactant side.

3Cl_{2 (g)} + 3H₂O_(l) → 5HCl_(aq) + HClO_{3 (aq)}

We cannot divide the coefficients to get any smaller whole numbers. Therefore, the equation is balanced.

Next, lets work on

Fe_(s) + O_{2 (g)} → Fe₂O_{3 (s)}

Since the Oxygen in Fe₂O₃ is an odd number and the reactant Oxygen is only available as O₂ , we can trade the two numbers as the coefficients to get the smallest whole amounts. This will produce a total of 6 oxygen on each side.

Fe_(s) + 3O_{2 (g)} → 2Fe₂O_{3 (s)}

Finally, notice that we now have 4 Iron on the product side. We can simply write this number as the coefficient for the atomic Iron reactant.

4Fe_(s) + 3O_{2 (g)} → 2Fe₂O_{3 (s)}

Now, both equations have been balanced.

Answer 2

Hello, Kristine,

Balancing equations can be challenging. Do you have a system that you've been using? I posted an approach that I use on Wyzant's Resources page:

https://www.wyzant.com/resources/files/729420/balancing_chemical_equations

It might prove helpful in these questions.

I often set up a spread sheet that allows me to change coefficients and have the element count update automatically on both sides. I tackle these problems by starting with the most complex molecule, which I decide should be HClO₃ in this case. We need at least one molecule, so pencil a coefficient of 1 for the HClO₃ . That means we need 1 Cl, but it only comes as Cl₂, so I quickly changed the coefficient to 2 molecules. Then I pencil in 1 for the Cl₂, and, so far, we're balanced. 2 HClO₃ molecules require 2 H and 3 O atoms. The source of O atoms is H₂O, so I put 3 for the H₂O coefficient. Next, I see that the 3 H₂O molecules provide 6 H atoms, that go to both the HCl and HClO₃. 2 H are used for the 2 HClO₃, leaving 4 H atoms. So I'll use them in the HCl by using a coefficient of 4 for HCl. But the H atoms come in twos (H₂O), so the sum on the right side must be even. If I change the HCl to 5, the H's are 6 on both sides. This leaves Cl. The products now have 5 HCl and 1 HClO₃. That makes a total of 6 Cl atoms. Cl comes in pairs (Cl₂), so we need a coefficient of 3. I think everything is now balanced. Please check the numbers. Painful, but fun when it works.

[3 ]CI2(g) + [ 3] H2O(I) ——-> [5 ] HCI(aq) + [1 ] HCIO3(aq)

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The second reaction is easier. Start with the most complex molecule, Fe₂O₃. Assign it a coefficient of 1 (in pencil). We need a minimum of 2 Fe atoms, so pencil a coefficient of 2 for Fe. Uh-oh, we need 3 O for the Fe₂O₃, but O only comes in pairs. So we need the change the coefficient for Fe₃O₃ to make it an even number of O's. Let's try 2. Now it's easy to figure out the coefficients for O₂ and Fe. To make 2 Fe₂O₃ molecules we need 4 Fe and 6 O atoms. Aha!

[4]Fe(s) + [3]O2(g) —-> [2]Fe2O3(s)

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It is easier once you've practiced these steps.

I hope this helps,

Bob