The document discusses several laws of chemical combinations and balancing chemical equations. It introduces the law of conservation of mass, law of definite proportions, law of multiple proportions, and law of reciprocal proportions. It also defines valency as the combining power of an element and discusses balancing chemical equations by ensuring the left and right sides match based on atom counts. An example calculation is provided to determine the mass of O2 needed to burn a given mass of sodium.

1. CHEMISTRY
CHEMICAL COMBINATIONS:
balancing chemical equations and laws
of chemical combination
Short revision series

2. Law of conservation of mass
 Matter can neither be created nor destroyed
during a chemical reaction but can be
converted from one form to another

3. Law of definite proportions
 States that all pure samples of a particular
chemical compound contain similar
elements combined in the same proportion
by mass

4. Law of multiple proportions
 The law of multiple proportions states that if
two elements A and B combine to form more
than one chemical compound, then the
various masses of one element A, which
combine separately with a fixed mass of the
other element B, are in simple multiple ratio.

5. Law of reciprocal proportions
 States that the masses of several elements A,
B, C, which combine separately with a fixed
mass of another element D, are the same as,
or simple multiples of, the masses in which
A, B, C, themselves combine with one
another.

6. Valency
 Is the combining power of an element.
 The number of moles of H atoms that will
combine with one mole of an element gives
the valency of that element.

7. Balancing chemical equations
 Cu + 2HCl  CuCl2 + H2
 all equations must adhere to law of
conservation of matter.
 The right hand side must match the left
hand side
 Do Atom count

8.  Eg balance
 HCl + CaCO3  CaCl + HO + CO

9. calculations
 Eg
 4 Na + O2  2Na2O
 what mass of O2 is needed to burn 5g of Na?

10. Solution to question
 4 moles of Na reacts with 1 mole of O2
 Mass of moles of Na = 4 x 23 = 92
 Mass of 1 mole of O2 = 2 x 16 = 32
 Therefore 92g of Na reacts with 32g of O2
 5g of Na reacts with 5 x 32 = 1.74g of O2
92