1. Bellwork Wednesday
How many atoms are in each of the following compounds?
H2SO4
Ca(NO3)2
(NH4)3PO4
C6H12O6
2 H2O
10 CO2
7 atoms!
9 atoms!
20 atoms!
24 atoms!
6 atoms!
30 atoms!

2. Part 1 CHEMICAL REACTIONS
CHEMISTRY I CHAPTER 11
Part 1 CHEMICAL REACTIONS

3. 11.1 Chemical reaction-
the changing of substances by the breaking of bonds in reactants and the formation of bonds in products.

4. Evidence of a chemical reaction:
1. Release of a gas
Zinc is added to hydrochloric acid producing hydrogen gas and solid zinc chloride.

5. 2. Color changes
two liquids are mixed
solid and a liquid mixed

6. 3. Formation of a precipitate
a precipitate is a solid product formed by the reaction of two aqueous solutions.
It is abbreviated ppt.
Aqueous sodium iodide and aqueous lead (II) nitrate produce solid lead (II) iodide and aqueous sodium nitrate.

7. 4. Changes in heat and light
all reactions either absorb or release energy
Sodium peroxide (yellow powder) and zinc powder (gray powder in the bowl) are combined
Water is then squirted in & the mixture ignites

8. Writing Chemical Equations:
yield
reactants  products
EXAMPLE:
Na Cl2  NaCl

9. Possible symbols in chemical equations:
+ plus
 yields
(s) solid
(l) liquid
(g) gas
(aq) aqueous
 equilibrium
N.R. no reaction
 heat is added
catalyst
 (light or heat) & catalysts are written above the yield sign

10. Examples: To what is the arrow pointing?
H2 (g) + O2 (g)  H2O (g)
MnO2
H2O2 (aq) H2O (l) + O2 (g) 
CaCO3 CaO (s) + CO2 (g)

11. Writing and Balancing Chemical Equations
Example:
Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.

12. sodium hydroxide + hydrogen
Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
Write the formulas of all reactants to the left of the arrow and all products to the right of the arrow.
Sodium + water
sodium hydroxide + hydrogen
Translate the equation and be sure the formulas are correct.
Na + H2O  NaOH + H2

13. Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
Once the formulas are correctly written, DO NOT change them. Use coefficients (numbers in front of the formulas), to balance the equation. DO NOT CHANGE THE SUBSCRIPTS!
_____Na + _____H2O  ____NaOH + _____H2

14. Begin balancing with an element that occurs only once on each side of the arrow.
Ex: Na 2 2 2
_____Na + _____H2O  ____NaOH + _____H2
When you are finished, you should have equal numbers of each element on either side of the equation Na H O 2 Na H O 2 4 4 2 2

15. To determine the number of atoms of a given element in one term of the equation, multiply the coefficient by the subscript of the element.
Ex: In the previous equation (below), how many hydrogen atoms are there? 4 2 2 2
____Na + _____H2O  ____NaOH + _____H2

16. Helpful Hints: Balance elements one at a time.
Balance polyatomic ions that appear on both sides of the equation as single units. (Ex: Count sulfate ions, not sulfur and oxygen separately)
Balance H and O last. Save the one that is in the most places for last…
Use Pencil!

17. Practice:
Balance the equation for the formation of magnesium nitride from its elements.
Mg2+
N3-
Mg3N2
____Mg ____N2  3
____Mg3N2

18. Balance the reaction of sodium metal with chlorine gas to form sodium chloride.
Na+
Cl-
NaCl
____Na ____Cl2  2
____NaCl 2

19. Coefficients are always whole numbers:
Sometimes though, fractions seem necessary.

20. Ex: NH O2  NO H2O
H can be balanced by placing a 2 in front of NH3 and a 3 in front of H2O. Then put a 2 in front of NO2 for nitrogen to balance.
_____NH _____O2  ____NO ____H2O 2 2 3

21. _____NH3 + _____O2  ____NO2 + ____H2O
7/2 2 3
Now all that is left to balance is the oxygen. There are 2 O on the reactant side and 7 on the product side. Our only source of oxygen is the O2. Any whole number we place in front of the O2 will result in an even number of atoms. The only way to balance the equation is to use a coefficient of 7/2.

22. _____NH3 + _____O2  ____NO2 + ____H2O
7/2 2 3
On a molecular level this makes no sense. You cannot have ½ of an O2 molecule. So…to get rid of the fraction, multiply all the coefficients by 2 (the denominator). ( ) 2
____NH _____O2  ____NO ____H2O 2
7/2 2 3 =
_____NH _____O2  ____NO ____H2O 4 7 4 6

23. Practice equations: ____H2 + ____O2  ____H2O
___NH3 + ____O2  ____NO ____H2O
___Ca + ___H2O  ___Ca(OH)2 + ___H2

24. ___ZnO + ___HCl  ___ZnCl2 + ___H2O
__NH4Cl +__Ca(OH)2  __NH3 + __H2O+__CaCl2
___ZnO + ___HCl  ___ZnCl2 + ___H2O
____K _____F2  _____KF
____C2H4 + ____O2  ____CO ____H2O

25. Bellwork Thursday Balance the following equations.
___Zn + ___HCl  ___ZnCl ___H2
 __Al2(SO4)3 + __Ca(OH)2  __Al(OH)3 + __CaSO4
  __K + __H2O  __KOH + __H2
  __CH4 + __O2  __CO __H2O