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Unit 8 – Chemical Equations and Reactions Notes p.1-2 These ones are in order!

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Format of Chemical Equations: 2H 2 + O 2 2H 2 O Reactants/Reagents

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Format of Chemical Equations: 2H 2 + O 2 2H 2 O Product(s)

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Format of Chemical Equations: 2H 2 + O 2 2H 2 O Yield(s)

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Format of Chemical Equations: 2H 2 + O 2 2H 2 O Subscript(s)

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Format of Chemical Equations: 2H 2 + O 2 2H 2 O Coefficients(s)

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Coefficients Large numbers in front Give how many molecules of each substance are in the equation Subscripts Small numbers after an element Give how many of each atom are needed in the formula Format of Chemical Equations:

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Now for some practice These are not in the notes but are essential skills to have

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What is the coefficient for … 2Al + 3CuSO 4 Al 2 (SO 4 ) 3 + 3Cu Aluminum: 2 Copper sulfate: 3 Aluminum sulfate: 1 (when there is no number written, it is 1) Copper: 3

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What is the coefficient for … 2H 2 O 2H 2 + O 2 Water: 2 Hydrogen gas: 2 Oxygen gas: 1

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How many ___ are there in ___ Mg(OH) 2 + 2HCl MgCl 2 + 2H 2 O How many H are in Mg(OH) 2 ? 2 (distribute the subscript through the parentheses) How many H are in HCl? 2 (the coefficient means there are 2 H) How many H are in the reactants? 4 total: 2 from HCl and 2 from Mg(OH) 2 How many H are in the products? 4 (all from H 2 O)

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Law of Conservation of mass Matter cannot be created nor destroyed, only change form Balancing Equations Must be the same number of each type of atom on the reactant side and the product side Even though a reaction has taken place, the mass at the beginning and the mass at end are the same. No mass (or matter) is gained or lost in the reaction.

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Never change SUBSCRIPTS in balancing!! Only change COEFFICIENTS!!!

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To Balance an Equation: 1. Total the number of each element for both sides 2. Check if each element is the same on both sides 3. Change the total on the inventory list so that the number of atoms matches. 4. Add a coefficient to the substance in the equation 5. Adjust the inventory list to reflect any changes in other atoms 6. Move on to each atom until the lists match for EVERY element

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Tricks: Balance polyatomic ions as a group If an element is present twice on one side, leave it for last If the list isn’t working because you have an odd number on one side and an even number on the other, try doubling everything Simplify: Check after balancing for any factor that could divide through all the coefficients

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EXAMPLE: WRITE A BALANCED EQUATION Chlorine + sodium bromide bromine + sodium chloride Cl 2 + NaBr Br 2 + NaCl Cl: 2Cl: 1Na: 1 Br: 1Br: 2 1. Total the number of each element for both sides Look at subscripts to tell how many are in the unbalanced equation All coefficients in an unbalanced equation are 1 (You might need to brush up on writing chemical formulas if you are only given the name.)

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EXAMPLE: WRITE A BALANCED EQUATION Chlorine + sodium bromide bromine + sodium chloride Cl 2 + NaBr Br 2 + NaCl Cl: 2≠Cl: 1 Na: 1=Na: 1 Br: 1 ≠ Br: 2 2. Check if each element is the same on both sides - Chlorine and bromine are unbalanced

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EXAMPLE: WRITE A BALANCED EQUATION Chlorine + sodium bromide bromine + sodium chloride Cl 2 + NaBr Br 2 + NaCl Cl: 2≠Cl: 1 2 Na: 1=Na: 1 Br: 1 ≠ Br: 2 3. Change the total on the inventory list so that the number of atoms matches. - I like to start with the first element on my inventory list. It doesn’t really matter which one you start with.

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EXAMPLE: WRITE A BALANCED EQUATION Chlorine + sodium bromide bromine + sodium chloride Cl 2 + NaBr Br 2 + 2 NaCl Cl: 2≠Cl: 1 2 Na: 1=Na: 1 Br: 1 ≠ Br: 2 4. Add a coefficient to the substance in the equation - 1 x coefficient = 2, therefore the coefficient is 2

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EXAMPLE: WRITE A BALANCED EQUATION Chlorine + sodium bromide bromine + sodium chloride Cl 2 + NaBr Br 2 + 2 NaCl Cl: 2 =Cl: 1 2 Na: 1 ≠ Na: 1 2 Br: 1 ≠ Br: 2 5. Adjust the inventory list to reflect any changes in other atoms - The new coefficient applies to both Na and Cl. According to the equation, there are now 2 Na in the products.

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EXAMPLE: WRITE A BALANCED EQUATION Chlorine + sodium bromide bromine + sodium chloride Cl 2 + 2 NaBr Br 2 + 2 NaCl Cl: 2 =Cl: 1 2 Na: 1 2 = Na: 1 2 Br: 1 2 =Br: 2 6. Move on to each atom until the lists match for EVERY element - Na is now unbalanced. The new coefficient in front of NaBr should be 2. That means there are now also 2 Br. The equation is balanced!

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + CaCl 2 AlCl 3 + CaSO 4 Al: 2Al: 1 SO 4 : 3SO 4 : 1Ca: 1 Cl: 2Cl: 3 1. Total the number of each element for both sides Look at subscripts to tell how many are in the unbalanced equation All coefficients in an unbalanced equation are 1 Trick: Because SO 4 is present as a polyatomic ion on both sides, it (sulfate) can be treated as one unit instead as S and O separately.

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + CaCl 2 AlCl 3 + CaSO 4 Al: 2 ≠ Al: 1 SO 4 : 3 ≠ SO 4 : 1 Ca: 1 =Ca: 1 Cl: 2 ≠ Cl: 3 2. Check if each element is the same on both sides - Aluminum, sulfate, and chlorine are unbalanced

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + CaCl 2 AlCl 3 + CaSO 4 Al: 2 = Al: 1 2 SO 4 : 3 ≠ SO 4 : 1 Ca: 1 =Ca: 1 Cl: 2 ≠ Cl: 3 3. Change the total on the inventory list so that the number of atoms matches. - I like to start with the first element on my inventory list. It doesn’t really matter which one you start with – but do one at a time!

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + CaCl 2 2 AlCl 3 + CaSO 4 Al: 2 = Al: 1 2 SO 4 : 3 ≠ SO 4 : 1 Ca: 1 =Ca: 1 Cl: 2 ≠ Cl: 3 4. Add a coefficient to the substance in the equation - 1 x coefficient = 2, therefore the new coefficient is 2. Now Al is balanced.

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + CaCl 2 2 AlCl 3 + CaSO 4 Al: 2 = Al: 1 2 SO 4 : 3 ≠ SO 4 : 1 Ca: 1 =Ca: 1 Cl: 2 ≠ Cl: 3 6 5. Adjust the inventory list to reflect any changes in other atoms - The new coefficient applies to both Al and Cl. According to the equation, there are now 6 Cl in the products (2 from coefficient x 3 from subscript).

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + 3 CaCl 2 2 AlCl 3 + CaSO 4 Al: 2 = Al: 1 2 SO 4 : 3 ≠ SO 4 : 1 Ca: 1 =Ca: 1 Cl: 2 6= Cl: 3 6 6. Move on to each atom until the lists match for EVERY element - Cl in the reactants is now unbalanced. There need to be 6. 2 x coefficient = 6, so the coefficient for CaCl 2 needs to be 3. Now Cl is balanced.

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + 3 CaCl 2 2 AlCl 3 + CaSO 4 Al: 2 = Al: 1 2 SO 4 : 3 ≠ SO 4 : 1 Ca: 1 3 =Ca: 1 3 Cl: 2 6= Cl: 3 6 6. Move on to each atom until the lists match for EVERY element - Changing the coefficient of CaCl 2 means there are now 3 Ca. To balance calcium we need to change the number in the products.

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EXAMPLE: WRITE A BALANCED EQUATION aluminum sulfate+ calcium chloride aluminum chloride +calcium sulfate Al 2 (SO 4 ) 3 + 3 CaCl 2 2 AlCl 3 + 3 CaSO 4 Al: 2 = Al: 1 2 SO 4 : 3 = SO 4 : 1 3 Ca: 1 3 =Ca: 1 3 Cl: 2 6= Cl: 3 6 6. Move on to each atom until the lists match for EVERY element - Calcium now balances! The new coefficient means there are three sulfates. Now sulfate balances!

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Practice Problems ___ C 3 H 8 + ___ O 2 ___ CO 2 + ___ H 2 O ___ NaBrO 3 ___ NaBr + ___ O 2 ___ NH 4 NO 3 ___ N 2 O + ___ H 2 O ___ H 2 + ___ CO + ___ O 2 ___ H 2 CO 3 ___ La 2 O 3 + ___ H 2 O ___ La(OH) 3 ___ NaI + ___ Pb(NO 3 ) 2 ___ NaNO 3 + ___PbI 2 15 3 4 2 2 3 1 111 132 2 121 1 12 Balance oxygen LAST! Balance oxygen LAST! Make sure to account for both N in the reactants. It was already balanced! Yay! Balance oxygen LAST! Treat the nitrate ion as one unit. HINTS

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Semester 1 Chemistry Review DAY 2 Formula Weight Find the formula weight of aluminum sulfate. Al 2 (SO 4 ) 3 Al - 2 x 26.98 = 53.96 S - 3 x 32.06 = 96.18.

Semester 1 Chemistry Review DAY 2 Formula Weight Find the formula weight of aluminum sulfate. Al 2 (SO 4 ) 3 Al - 2 x 26.98 = 53.96 S - 3 x 32.06 = 96.18.

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