1. Balancing Chemical Equations Chapter 8.1

2. Law of Conservation of Mass Mass cannot be created or destroyed only conserved. “What you start with is what you end with.”

3. Write the unbalanced equation. ◦Chemical equations written with reactants on right and products on the left. ◦Reactants  Products

4. Balance the equation. ◦Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation. Tip: Start by balancing an element that appears in only one reactant and product. ◦Balance chemical formulas by placing coefficients in front of them. Do not add subscripts, because this will change the formulas.

5. Coefficients are Multipliers for the entire compound. They are distributed just like they would be in a math class.

6. Worked Example Problem Tin oxide is heated with hydrogen gas to form tin metal and water vapor. The unbalanced equation: SnO 2 + H 2 → Sn + H 2 O

7. Now there are two Oxygen on each side of the equation Balance the equation In this case, there are two oxygen atoms on the left-hand side of the equation and only one on the right-hand side. Correct this by putting a coefficient of 2 in front of water: SnO 2 + H 2 → Sn + H 2 O SnO 2 + H 2 → Sn + 2 H 2 O

8. Now there are 4 hydrogens on each side of the quation. This puts the hydrogen atoms out of balance. Now there are two hydrogen atoms on the left and four hydrogen atoms on the right. SnO 2 + H 2 → Sn + 2 H 2 O To get four hydrogen atoms on the right, add a coefficient of 2 for the hydrogen gas. SnO 2 + 2 H 2 → Sn + 2 H 2 O

9. Great work! The equation is now balanced. Be sure to double-check your math! SnO 2 + 2 H 2 → Sn + 2 H 2 O Each side of the equation has 1 atom of Sn, 2 atoms of O, and 4 atoms of H.

10. Remember Remember, coefficients are multipliers, so if we write 2 H 2 O it denotes 2x2=4 hydrogen atoms and 2x1=2 oxygen atoms.

11. Let’s see another… Sometimes the equations doesn’t balance perfectly the first attempt. Al + O 2  Al 2 O 3 Adding a 2 in front of the Al will keep the Oxygens out of balance. Al + 3 O 2  2 Al 2 O 3 Adding a 3 on the left O and a 2 in front of the right O will make 6 Oxygens on each side. 4 Al + 3 O 2  2 Al 2 O 3 Now balancing the Aluminums was simple.

12. 2.5 becomes 5 and everything else remains balanced after doubling! N 2 + O 2 + H 2 O  HNO 3 Currently: 2 N on right, 1 N on left. 3 O on right, 3 O on left. 2 H on right, 1 H on left. Let’s balance H. N 2 + O 2 + H 2 O  2 HNO 3 2 H on right, 2 H on left. Now the Nitrogens are balanced 2 on right, 2 on left. 3 O on left and 6 O on right. Let’s balance O. N 2 + O 2 + H 2 O  2 HNO 3 Oh no! If we try to balance the Oxygens the other elements become unbalanced! Adding a 2in front of the O will give only 5 Oxygen on the left, adding a 3 will give 7. 2.5 would be perfect! At this point, double everything to make 2.5 work: 2 N 2 + 5 O 2 + 2 H 2 O  4 HNO 3

13. Now complete Section Review 8.1 Reminders: ◦Coefficients get distributed ◦You may run into problems on your first attempt. ◦If a decimal would work, double all the coefficients!