Catalyst – November 6(2), 2009  Balance the following equations: 1. Na + H 2 O → H 2 + NaOH 2. ZnS + O 2 → ZnO + SO 2 Please take out HW to get stamped!

Today’s Agenda  Catalyst  Go over HW  More Balancing Reactions, but with a little sauce  Differentiated Balance HW  Exit Question

Today’s Objectives  SWBAT balance chemical equations with ease!

Remember…  Why do we have to balance chemical reactions?  Law of Conservation of Matter!

Balancing with some hot sauce… Key Point #1: When balancing equations that involve polyatomic ions, keep them together! What if you have some polyatomic ions? WHAAAAAAAAT will you do??

Work it out… H 3 PO 4 + ____ KOH  ____ K 3 PO 4 + ____ H 2 O  Which polyatomics are we using?  PO 4 – phosphate  OH – hydroxide  Where is the OH on the other side?  It’s okay – rewrite H 2 O as HOH  Trust me!

Balance it like its Hotttt H 3 PO 4 + ____ KOH  ____ K 3 PO 4 + ____ HOH

Your turn  HCl + NaOH  NaCl + H 2 O  NaOH + H 2 CO 3  Na 2 CO 3 + H 2 O  Al(OH) 3 + H 2 CO 3  Al 2 (CO 3 ) 3 + H 2 O

Balancing with diatomics… Key Point #2: Sometimes with diatomics, you must be a little tricky.  C 4 H 10 + O 2  CO 2 + H 2 O

Diatomics  If you reach the point in a balancing problem where everything is balanced except the diatomic atom...  You can only get even numbers with diatomics…  2 O 2 = 4  3 O 2 = 6  And sometimes the other side is odd  Use a decimal (fraction), then multiply by 2

Work it out… C 4 H 10 + O 2  CO 2 + H 2 O

Your Turn…  C 6 H 6 + O 2  CO 2 + H 2 O  C 10 H 22 + O 2  CO 2 + H 2 O

Practice, Practice, Practice…  You must complete TWO sections: Easy, Medium, Hard, or Impossible  There are 12 problems in each section (2 sets of 6)  Each EXTRA section is worth 20 BONUS POINTS

Exit Question  Balance the following equation: Si(OH) 4 + NaBr  SiBr 4 + NaOH