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} Ch 9 section 1 Performance Goal:

} Writing a Chemical Equation: ◦ Must be balanced and include proper Need books and notes symbols

} Background knowledge: what do you already } What is the difference between: know? vocab? } Word equation } Skeleton equation } Check that you can answer the essential } Chemical equation questions as you study

} #1 is signs of a chemical reaction! } Read and discuss in small groups for 2 minutes } Word- no formulas is a BALANCED equation reactants, products and how much of } Skeleton- formulas but not balanced each

} Chemical- formulas AND it’s balanced (visual page 285)

} Which one is the performance goal?

} We must start and end with the same stuff. WHY? P. 288 Besides formulas: See page 283 vocab and symbols used } Conservation of Mass In your notes ADD this symbol: ∆

When heating a reaction– place the symbol ∆ above the arrow } subscripts tell you ◦ how many atoms of a particular ◦ When balancing: element are in a compound. ◦ You add coefficients in front of the compounds (chapter 7 and 8)

◦ but you may not change the subscripts. } The coefficient tells you ◦ about the quantity of that compound in ◦ WHY???? a reaction (chapter 9)

Al (s) + O 2 (g) ‰‰‰ Al 2O3 (s) 1) Polyatomics Use LCM idea: 2 and 3 of : use total of 6 that are on BOTH sides- balance these as a unit Al (s) + 3 O 2 (g) ‰‰‰ 2 Al 2O3 (s) FYI: 4Al (s) + 3 O 2 (g) ‰‰‰ 2 Al 2O3 (s) Time saver- instead of doing individual atoms making up the poly, you do the whole unit take the coefficient TIMES subscript so there are 6 O’s on the left and the right 2) Singletons (“solo”acts) 3) Balance other atoms

Balance atoms (except H and O) that appear only • 4) Usually save H or O for last (they may ONCE on EACH side appear in several compounds) and are then easier if last

• There is no one right way every time • (your book has a general guide on p. 286-288)

} Ag NO Cu ‰ Cu NO Ag } AgNO 3 + Cu ‰ Cu(NO 3)2 + Ag 3 3 } 1 1 1 1 2 1 } Balance polys first

} 2 AgNO 3 + Cu ‰ Cu(NO 3)2 + Ag } 2 AgNO 3 + Cu ‰ Cu(NO 3)2 + Ag } 2 2 1 1 2 1 } Then Ag falls out nicely } 2 AgNO 3 + Cu ‰ Cu(NO 3)2 + 2 Ag } 2 AgNO 3 + Cu ‰ Cu(NO 3)2 + 2 Ag } 2 2 1 1 2 2 ◦ No need for N or O if you do the poly as a whole! } Al 2O3 + CuCl 2 ‰ CuO + AlCl 3 } Al 2O3 + CuCl 2 ‰ CuO + 2AlCl 3

} Then balance Cl

} This type you can start anywhere; } Al 2O3 + 3 CuCl 2 ‰ CuO + 2 AlCl 3 (every element is a “solo act”!) } Then Cu

} Al 2O3 + 3 CuCl 2 ‰ 3CuO + 2 AlCl 3 } I used Al first for this ex. } O is already done!

PbSO 4 + AlF 3 ‰ PbF 2 + Al 2(SO 4)3 } Do sulfate first } 3 PbSO 4 + AlF 3 ‰ PbF 2 + Al 2(SO 4)3

Where do you start? ◦ Then Pb (or Al, F, doesn’t matter!) } 3 PbSO 4 + AlF 3 ‰ 3 PbF 2 + Al 2(SO 4)3

} 3 PbSO 4 + 2 AlF 3 ‰ 3 PbF 2 + Al 2(SO 4)3

Last check Al } There is a category of equations called } This is the order of elements to balance COMBUSTION } Works EVERY time! } It is a C and H compound burning in oxygen st (remember the element oxygen is a diatomic!) } 1 , using CO 2 } then H using the water } then finish with O on the left with the diatomic oxygen } PRODUCTS are CO 2 and H 2O

} EX: __ CH 4 + __ O 2 ‰ ___ CO 2 + ___ H 2O

} __ CH 4 + __ O 2 ‰ ___ CO 2 + _2_ H 2O

___ CH 4 + __ O 2 ‰ ___ CO 2 + ___ H 2O } Oxygen: Have 4 oxygen on the right

Carbons are equal, move to H } (2 from CO 2 and 2 from H 2O) 4 H on left, Need 2 water on right ◦ put a 2 for oxygen on the left

___ CH 4 + __ O 2 ‰ ___ CO 2 + _ 2_ H 2O } CH + 2 O ‰ CO + 2 H O 4 H’s on left and now 4 H’s on right 4 2 2 2 ___ C H + __ O ‰ 4 CO + 5 H O EX: ___ C 4H10 + __ O 2 ‰ __ CO 2 + __ H 2O 4 10 2 2 2 On right side: 8 O + 5 O = 13 O 4 C’s need 4 CO ’s 2 Tip: Odd # of O on the right? Put it over 2, (Why? X times 2 = 13, Solving for X) ___ C 4H10 + __ O 2 ‰ 4 CO 2 + __ H 2O then double all coefficients.

10 H’s need 5 waters ___ C 4H10 + 13/2 O2 ‰ 4 CO 2 + 5 H2O

2 C4H10 + 13 O2 ‰ 8 CO 2 + 10 H2O ___ C 4H10 + __ O 2 ‰ 4 CO 2 + 5 H2O Why double- no half molecules!

EX: __C 3H8O3 + __ O 2 ‰ __CO 2 + __ H 2O __C 3H8O3 + __ O 2 ‰ 3 CO 2 + 4 H2O Right side: 6 O + 4 O = 10 O

__C 3H8O3 + __ O 2 ‰ 3 CO 2 + __ H 2O Left : 3 are already in formula,

__C 3H8O3 + __ O 2 ‰ 3 CO 2 + 4 H2O

Need 10-3 = 7 needed using the O 2 Watch out here molecule

X times 2 = 7 X = ? (7/2) __C 3H8O3 + 7/2 O2 ‰ 3 CO 2 + 4 H2O

Last DOUBLE all coefficients: } On to the packet of 50 balancing equations

} Do not put 1’s in the blanks 2 C3H8O3 + 7 O2 ‰ 6 CO 2 + 8 H2O } 1’s are understood

} If an equation is already balanced (all 1’s) simply mark it so- check mark, or write balanced (bal.) or write done for example

Sodium phosphate + (III) ‰‰‰ } phosphate + iron (III) oxide ‰‰‰ } sodium oxide + iron (III) phosphate sodium oxide + iron (III) phosphate } Na 3PO 4 + Fe 2O3 ‰‰‰

} Skeleton first: NEED FORMULAS!! } Na 2O + FePO 4 } Poly is equal on both sides; } Sodium phosphate + iron (III) oxide ‰‰‰ sodium } Your choice, either Fe or Na first oxide + iron (III) phosphate

} } Na 3PO 4 + Fe 2O3 ‰‰‰ Na 2O + FePO 4 2 Na 3PO 4 + Fe 2O3 ‰‰‰ 3 Na 2O + 2 FePO 4