Download Balancing Reactions

Chapter 3.7: Chemical Reactions (Balancing)
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Balancing Chemical Equations
o Why do chemical equations need to be balanced? Conservation of Mass!
o Helpful hints:
1. Aim for whole numbers by increasing coefficients
2. Balance elements appearing in only one species on either side of reaction first
• e.g. in C4H10 + O2 → CO2 + H2O, balance C & H first:
C4H10 + O2 → 4 CO2 + 5 H2O
Now you can balance O2:
C4H10 + 6½ O2 → 4 CO2 + 5 H2O
Double the coefficients to eliminate the fraction:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
3. Balance individual elements last—any coefficient necessary can be used to balance these without
affecting any other species. (as with O2 in above example)
4. Balance elements with an even subscript on one side and an odd subscript on the other by switching
these subscripts as the coefficients
• e.g. Fe + O2 → Fe2O3, balance O first:
Fe + 3 O2 → 2 Fe2O3 (notice how the subscripts on O became the coefficients)
Now you can balance Fe as in #2 above:
4 Fe + 3 O2 → 2 Fe2O3
5. Balance elements with high subscripts on one side first, if possible (esp. double replacement)
• e.g. BaCl2 (aq) + Na3PO4 (aq) → NaCl (aq) + Ba3(PO4)2 (s), start with Ba3PO4:
We need 3 Ba and 2 PO4 on the reactant sde:
3 BaCl2 (aq) + 2 Na3PO4 (aq) → NaCl (aq) + Ba3(PO4)2 (s)
Now it’s easy to see that we need 6 Na and 6 Cl in the products:
3 BaCl2 (aq) + 2 Na3PO4 (aq) → 6 NaCl (aq) + Ba3(PO4)2 (s)
6. If an element appears in one species on one side and multiple species on the other side, increasing
the coefficient on the single species side one at a time will often help lead to the solution. Be aware
that you may also need to consider whether you need to have an even number or multiple of some
number.
• e.g. Cu + HNO3 → Cu(NO3)2 + NO2 + H2O (note: this is similar to one of the homework
problems, but the products are slightly different—the process is the same)
Here we have N in two products. Since H is only in H2O in the products, HNO3 must have an
even coefficient. Start with 2:
Cu + 2 HNO3 → Cu(NO3)2 + NO2 + H2O
This gives 2 N in the reactants, but we have 3 N in the products, so we need to have at least that
many N in the reactants, so increase the coefficient on HNO3 to 4:
Cu + 4 HNO3 → Cu(NO3)2 + NO2 + H2O
Now we need to increase the number of N on the product side. Increasing the coefficient on
Cu(NO3)2 would make 5 N, but a 2 on NO2 would be right:
Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + H2O
Now all we need is to balance H:
Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + 2 H2O, et voila!
7. Reduce coefficients by Greatest Common Factor, if needed
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Homework #3-4: Problems pg. 106 #3.57, 3.58, 3.60
For the following reactions please write out all chemical formulas and balance equations:
a) Dicarbon dihydride (ethyne) + Oxygen → Carbon dioxide + Water
b) Iron (III) chloride + Ammonium hydroxide → Iron(III) hydroxide + Ammonium chloride
c) Phosphorus pentachloride + water → Hydrochloric acid + Phosphoric acid
d) Lead(II) nitrate → Lead(II) oxide + Nitrogen dioxide + Oxygen