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THIS
IS
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With
Host...
Your
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100 100 100 100 100 100
200 200 200 200 200 200
300 300 300 300 300 300
400 400 400 400 400 400
500 500 500 500 500 500
Mole-Mole Mass-Mass
Limiting & Excess
Yield Equations Rxn Types Predicting Products
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A 100
2 NH3 N2 + 3 H2
How many moles of nitrogen are produced when 1.5 moles of
ammonia decomposes?
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A 100
1.5 = X 0.75 mol
2 1
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A 200
4 Al + 3 O2 2 Al2O3
How many moles of aluminum are needed to
completely react with 2 moles of oxygen?
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A 200
X = 2 2.67 mol
4 3
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A 300
C3H8 + 5 O2 3 CO2 + 4 H2O
How many grams of water are produced when 45.0 grams of C3H8 combusts completely?
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A 300
45.0 g C3H8 x 1 mol = 1.023 mol
44 g
1.023 = X
1 4
4.09 mol H2O x 18 g = 73.6 g 1 mol
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A 400
C3H8 + 5 O2 3 CO2 + 4 H2O
When 150 grams of oxygen react, how many grams of carbon
dioxide will be produced?
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A 400
150 g O2 x 1 mol = 4.7 mol
32 g
4.7 = X
5 3
2.813 mol CO2 x 44 g = 123.75 g 1 mol
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A 500
C3H8 + 5 O2 3 CO2 + 4 H2O
If 36 grams of water are produced, how much carbon dioxide was also produced?
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A 500
36 g H2O x 1 mol = 2 mol
18 g
2 = X
4 3
1.5 mol CO2 x 44 g = 66 g 1 mol
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B 100
Define Limiting Reagent and Excess
Reagent
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B 100
Limiting Reagent – substance that gets completely used up in a reaction (limiting the amount of product formed)
Excess Reagent – substance that is not completely used up in a reaction (some is left over)
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B 200
How do you determine which
reactant is the limiting reagent?
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B 200
Find the amount of product that can be formed for each of the reactants. The one that can produce the least amount of product is the limiting reagent
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B 300
4 Al + 3 O2 2 Al2O3
Which is the limiting reagent if 7 moles of aluminum reacts
with 5 moles of oxygen?
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B 300
Al 7 = X 3.5 mol
4 2
O2 5 = X 3.3 mol
3 2
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B 400
4 Al + 3 O2 2 Al2O3
What is the maximum amount of aluminum oxide that can be
produced if 7 moles of aluminum reacts with 5 moles
of oxygen?
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B 400
Al 7 = X 3.5 mol 4 2
O2 5 = X 3.3 mol 3 2
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B 500
4 Fe + 3 O2 Fe2O3
What is the limiting reagent and what quantity of Fe2O3 can be formed if 74.55 g of iron reacts with 27.65 g of
oxygen?
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B 500
Fe 74.55 g x 1 mol = 1.33 mol 1.33 = X 0.3325 mol x 160 g = 54.25 g
56 g 4 1 1 mol
O2 27.65 g x 1 mol = 0.864 mol 0.864 = X 0.288 mol x 160 g = 46.08 g
32 g 3 1 1 mol
O2 / 46.08 g
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C 100
When you use a balanced equation to predict the
number of grams of product in a reaction, you get the
______________.
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C 100
Theoretical Yield
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C 200
In an experiment, students determined the theoretical yield to be 20 grams. Their actual yield was 15 grams.
What was their percent yield?
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C 200
15 x 100 = 75%
20
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C 300
In an experiment, students determined the theoretical yield to be 12 grams. Their
percent yield was 90 percent. What was their actual yield?
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C 300
X x 100 = 90% 12
10.8 g
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C 400
Is it possible to have a percent yield greater than 100%
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C 400
No, the maximum amount of product that can actually be produced can never be greater than the calculated theoretical yield
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C 500
4 NH3 + 5 O2 4 NO + 6 H2O
The reaction of 0.68 g of NH3 with excess O2 according to the following reaction yields 0.98 g
of NO. What is the percent yield?
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C 500
0.68 g NH3 x 1 mol = 0.04 0.04 = X 0.04 mol NO x 30 g = 1.2 g
17 g 4 4 1 mol
0.98 x 100 = 81.7%
1.2
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D 100
Name the reactantsMnO2 + Al → Al2O3 + Mn
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D 100
Manganese (IV) oxide
&
Aluminum
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D 200
Balance this equation
MnO2 + Al → Al2O3 + Mn
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D 200
3MnO2 + 4Al → 2Al2O3 + 3Mn
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D 300
What is meant by a balanced equation
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D 300
Obeying the law of conservation of matter: same
number of atoms of each element in reactants and
products
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D 400
Write a complete chemical equation for
aluminum sulfate + sodium hydroxide → aluminum hydroxide + sodium sulfate
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D 400
Al2(SO4)3 +6NaOH 2Al(OH)3 + 3Na2SO4
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D 500
Write the combustion reaction for
C2H4ç
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D 500
C2H4 + 3O2 2H2O + 2CO2
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E 100
Reaction Type?
C10H8 + 12O2 10CO2 + 4H2O
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E 100
Combustion
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E 200
Reaction Type?
2H2O 2H2 + O2
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E 200
Decomposition
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E 300
Reaction Type?
Mg + 2H2O Mg(OH)2 + H2
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E 300
Single Replacement
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E 400
Reaction Type?
8 Fe + S8 8 FeS
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E 400
Combination
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E 500
Reaction Type?
Pb(NO3)2 + 2KI PbI2 + 2KNO3
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E 500
Double Replacement
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F 100
Al + SnCl2 →
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F 100
2Al + 3SnCl2 → 3Sn + 2AlCl3
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F 200
Cl2 + NaBr →
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F 200
Cl2 + 2NaBr → 2NaCl + Br2
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F 300
Pb(NO3)2 + K2CrO4 →
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F 300
Pb(NO3)2 + K2CrO4 → PbCrO4 + 2KNO3
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F 400
Cu + ZnSO4
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F 400
No Reaction!
Copper is less reactive than zinc
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F 500
A piece of magnesium is dropped into hydrochloric acid (aqueous hydrogen
chloride)
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F 500
Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq)
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Final Jeopardy
Please record your wager.
Click on screen to begin
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Click on screen to continue
What is the limiting reagent and what amount of product can be formed if 44 grams of aluminum and 44 grams of fluorine are reacted?
If 60 grams are actually produced in the lab, what is the percent yield?
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Click on screen to continue
2 Al + 3 F2 2 AlF3
Al 44 g x 1 mol = 1.63 mol 1.63 = X 1.63 mol AlF3 x 84 g = 136.9 g 27 g 2 2 1 mol
F2 44 g x 1 mol = 1.16 mol 1.16 = X 0.77 mol AlF3 x 84 g = 64.8 g 38 g 3 2 1 mol
F2 / 64.8 g 60 x 100 = 92.6% 64.8
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Thank You for Playing Jeopardy!
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