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!

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

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

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

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

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.

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

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

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

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)

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

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

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

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

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

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!

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

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

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

Ex: NH3 + O2  NO2 + 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. _____NH3 + _____O2  ____NO2 + ____H2O 2 2 3

_____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.

_____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 ____NH3 + _____O2  ____NO2 + ____H2O 2 7/2 2 3 = _____NH3 + _____O2  ____NO2 + ____H2O 4 7 4 6

Practice equations: ____H2 + ____O2  ____H2O ___NH3 + ____O2  ____NO2 + ____H2O ___Ca + ___H2O  ___Ca(OH)2 + ___H2

___ZnO + ___HCl  ___ZnCl2 + ___H2O __NH4Cl +__Ca(OH)2  __NH3 + __H2O+__CaCl2 ___ZnO + ___HCl  ___ZnCl2 + ___H2O ____K + _____F2  _____KF ____C2H4 + ____O2  ____CO2 + ____H2O

Bellwork Thursday Balance the following equations. ___Zn + ___HCl  ___ZnCl2 + ___H2  __Al2(SO4)3 + __Ca(OH)2  __Al(OH)3 + __CaSO4   __K + __H2O  __KOH + __H2   __CH4 + __O2  __CO2 + __H2O