# Chapter 11 Stoichiometry.

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Chapter 11 Stoichiometry

½ c butter ½ c brown sugar ½ c granulated sugar 1 egg ½ tsp vanilla 1 cup flour ½ tsp salt ½ tsp baking soda ½ c chopped walnuts ½ c chocolate chips What is the product of this recipe? How would you adjust the ingredients in the recipe to make 7 dozen cookies? What would you do if you only had ¼ c butter?

Stoichiometry Mass-mass relationships
Quantitative study of chemical reactions

Balanced Equations give us:
the relationship between the reactants and the products the ratios in which they combine The coefficients represent the number of moles 2 H2 + 1 O2  2 H20 2 moles of hydrogen react w/ 1 mole of oxygen to form 2 moles of water

Law of Conservation of Mass
Basis behind stoichiometry Mass of reactants= mass of product Ex. 4Fe + 3O2  2 Fe2O3 4 moles of iron react w/ 3 moles of oxygen to form 2 moles of rust Convert moles to grams 4 moles Fe 55.8 g Fe = g Fe 1mole Fe Total mass of reactants: g 3 moles O2 32.0 g O2 = g O2 1 mole O2

Mass of reactants= mass of products
2 moles Fe2O3 158.7 g Fe2O3 = g Fe2O3 (mass of products) 1 mole Fe2O3 Mass of reactants= mass of products

Trail Mix In order to make trail mix, you need raisins, peanuts, and M&M's. What makes it good trail mix is having the right ratios! In the bags there are 1 dozen raisins (R), 2 dozen peanuts (P), and 3 dozen M&M's (MM). This is a perfect combination to make 1 package of trail mix (TM). If we were to write an equation to represent this recipe it would look like this: 1 dozen raisins + 2 dozen peanuts + 3 dozen M&M's = 1 trail mix or 1 R + 2 P + 3 MM  1 TM

1 R + 2 P + 3 MM1 TM With this in mind, what if you only had ½ the contents of your bags… How many dozen peanuts would you need to mix them with 1/2 dozen raisins? By looking at the equation, you can come up with a conversion of 2 dozen peanuts for every 1 dozen raisins. SO… ½ dozen raisins x 2 dozen peanuts = 1 dozen peanuts needed 1 dozen raisins (or ½ your bag!) (Just like halving a recipe!)

1 R + 2 P + 3 MM1 TM How many dozen M&M's would you need to combine them with 6 dozen peanuts? 3 dozen MM 6 dozen peanut 2 dozen peanuts 9 dozen m &m’s

1 R + 2 P + 3 MM1 TM How many dozen raisins would you need to mix with 4 dozen peanuts? 4 dozen peanut 1 dozen raisin 2 dozen peanut = 2 dozen raisins

1 R + 2 P + 3 MM1 TM How many dozen peanuts would you need to mix them with 18 M&M's?(careful!) 2 dozen peanuts 18 mm’s 1 dozen mm 3 dozen mm’s 12 mm’s 1 dozen peanuts

Chemists use ratios such as these ALL the time in the laboratory
Chemists use ratios such as these ALL the time in the laboratory. Instead of dozens, they use moles – something we've already looked at. Just as in the equation above, the coefficients in front of the compounds provide the "mole ratios" the chemists need.

Mole Ratios Ratio between the numbers of moles of any 2 substances in a balanced equation You can have many different ratio scenarios for each equation

2 Al + 3 Cl2  2 AlCl3 Ratio of Al to Cl 2 moles Al 3 moles Cl
Ratio of Cl to Al 3 moles Cl 2 moles Al Ratio of Cl to product 3 moles Cl 2 moles AlCl3 Ratio of product to Cl 2 moles AlCl3 Ratio of Al to product 2 moles Al Ratio of product to Al 2 moles AlCl3

How much of 1 reactant is needed to combine w/ a given amount of another reactant How much product will be produced w/ a given amount of reactant How much reactant is needed to produce a given amount of product

Steps in Stoichiometry
Write the balanced equation Write the given information under the equation (molar mass, # of moles, grams of known) Convert grams of known substance to moles using molar mass

Determine the mole ratio using the coefficients from balanced equation
moles of unknown moles of known 5. Convert the moles of unknown to grams using molar mass

known moles (molar mass unknown) g unknown Known mass g (from problem)
unknown moles 1 mole known g known (molar mass known) known moles ( coef. from eqn.) 1 mole unknown

How many grams of AlCl3 are produced if 3
How many grams of AlCl3 are produced if 3.0g of Cl2 react with excess aluminum? 1.) Balanced Equation: 2 Al Cl2  2 AlCl3 3.0 g 3 mole Cl2 71.0 g/mole ? g 2 moles AlCl3 133.5 g/mole 2.) Write in known info

3. Convert known grams to moles, put in mole ratio (from coefficients) and convert unknown moles to grams. 3.0g Cl2 2 moles AlCl3 133.5 g AlCl3 1 mole Cl2 3 moles Cl2 1 mole AlCl3 71.0 g Cl2 (Mass known) (Molar mass known) (Mole Ratio) (molar mass unknown)

3. Multiply the top line and divide by the bottom (making sure to use parentheses correctly)
(3.0 x 2 x ) (71.0 x 3) = g of AlCl3 formed

Stoichiometry of Shuttle Launch
NASA TV (Discovery Last Launch) You Tube (STS 119)

Percent Yield Quantities found in stoichiometry are theoretical or predicted amounts (the amount you are supposed to get if all the conditions are perfect) During actual experimentation, you will come up w/ less than expected % Yield = actual amt (experiment) x 100 Theoretical amt (calculated)

You calculated that you should get 82. 2g of NaCl in a reaction
You calculated that you should get 82.2g of NaCl in a reaction. When you performed the experiment you produced only 30.7 g NaCl. What is your % yield? % Yield = x 100 82.2 37.3%

Limiting Reactant (reagent)
Reactant that is totally consumed during a reaction Limits the extent of a reaction Determines the amount of product that can be formed Once it is gone, the reaction stops

Excess Reactant (reagent)
Reactant that remains after the reaction stops

Limiting Reactant Problems
+ 8 car bodies 16 excess tires + 48 tires 8 cars +

Steps: Balance equation Using known amounts of each reactant, solve for mass of product (do 2 separate stoichiometry problems) The one that produces the least amount of product is the limiting reactant.

Balance equation and write in known info
75 g of CaO react with 30.0 g HCl to produce calcium chloride and water. What is the limiting reactant? How much water is formed? Balance equation and write in known info CaO HCl  CaCl H20 75 g 1 mole 56.1 g/mole 30.0 g 2 mole 36.5 g/mole ? g 1 mole 18.0 g/mole

Do 2 stoichiometry problems, w/ each reactant as a known and the unknown being water
1 mole H2O 18.0 g H2O 75 g CaO 1 mole CaO 1 mole CaO 1 mole H2O 56.1 g CaO =24 g H2O

Do 2 stoichiometry problems, w/ each reactant as a known and the unknown being water
1 mole H2O 18.0 g H2O 30.0 g HCl 1 mole HCl 2 mole HCl 1 mole H2O 36.5 g HCl 7.40 g H2O

HCl is the limiting reactant, producing 7.40 g of water
CaO is in excess (meaning some will be left over after the reaction)

Finding mass of excess reactant that remains:
One can find the mass of excess reactants that will remain by doing another stoichiometry problem using your limiting reactant as your known and the excess reactant as your unknown Mass excess starting- mass excess used (stoich. Calc)= mass of left over excess react.

Using HCl as the known and CaO as the unknown, calc. the amt
Using HCl as the known and CaO as the unknown, calc. the amt. Of CaO used. 30.0 g HCl mole HCl 1 mole CaO g CaO 36.5 g HCl 2 mole HCl mole CaO =23.1 CaO used 75 g CaO – 23.1g used = 51.9 g leftover

N H2  2 NH3 If you have g of N2 and 25.0 g of H2, which is your limiting reactant in the reaction? How many grams of NH3 can be produced from that limiting reactant? Calculate the mass of excess reactant that remains after the reaction is complete.

Ch. 11 Test Review (stoich song)
Vocabulary stoichiometry Limiting reactant Excess reactant Mole ratio Actual yield (experimental) Theoretical yield (calculated) % yield

Practical application of stoichiometry
Calculations mole ratios Law of conservation of mass Stoichiometry Limiting reactant problem % yield Essay Practical application of stoichiometry

4HCl + O2  2 H2O + 2 Cl2 Interpret it in terms of moles
What is the mole ratio between water and oxygen? Show that mass is conserved

_ Cu + _ AgNO3  _ Cu(NO3)2 + _ Ag
Balance the equation How many grams of copper are needed to react with 12.0 g of AgNO3?

_ Al + _ Cl2  __AlCl3 Balance
If you begin with 3.2 g of aluminum and 5.4g of chlorine what is the limiting reactant?

__Mg+ _ HCl  _MgCl2 + _ H2 Balance
If you reacted 10.0g of magnesium with excess hydrochloric acid. How many grams of MgCl2 will be formed? After actually reacting this you formed 29.5 g of MgCl2, what is the percent yield?

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