Problem L35

[Jaden 14B 3F]

I am having trouble balancing the third equation in problem L35. Has anyone successfully done so and can provide help? Thank you.

[Timothy Li 3A]

Note: For the 3rd equation, the reactant should be Fe3Br8 and not FeBr2. That is a mistake in the textbook.

I would start by trying to balance the element that occurs the least. On the left of the equation, there are 8 moles of Br in Fe3Br8 so we know that there must be 8 moles of Br on the right side of the equation. We would thus get 8 NaBr on the right and Br is now balanced. Now the Na moles are imbalanced however, as we have 8 moles of Na on the right and only 1 mole of Na2 on the left. To balance that, we put a 4 in front of Na2CO3 on the left to get 4 Na2CO3. Now we have just C and O to balance. From 4 Na2CO3, we know that there are 4 moles of C and 12 moles of O. All we would need to do is put a 4 in front of CO2 on the right of the equation to balance those out. The final formula is

Fe3Br8 + 4 Na2CO3 --> 8 NaBr + 4 CO2 + Fe3O4