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Mole-Mole Mass-Mass

Limiting & Excess

Yield Equations Rxn Types Predicting Products

A 100

2 NH3 N2 + 3 H2

How many moles of nitrogen are produced when 1.5 moles of

ammonia decomposes?

A 100

1.5 = X 0.75 mol

2 1

A 200

4 Al + 3 O2 2 Al2O3

How many moles of aluminum are needed to

completely react with 2 moles of oxygen?

A 200

X = 2 2.67 mol

4 3

A 300

C3H8 + 5 O2 3 CO2 + 4 H2O

How many grams of water are produced when 45.0 grams of C3H8 combusts completely?

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

A 400

C3H8 + 5 O2 3 CO2 + 4 H2O

When 150 grams of oxygen react, how many grams of carbon

dioxide will be produced?

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

A 500

C3H8 + 5 O2 3 CO2 + 4 H2O

If 36 grams of water are produced, how much carbon dioxide was also produced?

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

B 100

Define Limiting Reagent and Excess

Reagent

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)

B 200

How do you determine which

reactant is the limiting reagent?

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

B 300

4 Al + 3 O2 2 Al2O3

Which is the limiting reagent if 7 moles of aluminum reacts

with 5 moles of oxygen?

B 300

Al 7 = X 3.5 mol

4 2

O2 5 = X 3.3 mol

3 2

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?

B 400

Al 7 = X 3.5 mol 4 2

O2 5 = X 3.3 mol 3 2

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?

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

C 100

When you use a balanced equation to predict the

number of grams of product in a reaction, you get the

______________.

C 100

Theoretical Yield

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?

C 200

15 x 100 = 75%

20

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?

C 300

X x 100 = 90% 12

10.8 g

C 400

Is it possible to have a percent yield greater than 100%

C 400

No, the maximum amount of product that can actually be produced can never be greater than the calculated theoretical yield

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?

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

D 100

Name the reactantsMnO2 + Al → Al2O3 + Mn

D 100

Manganese (IV) oxide

&

Aluminum

D 200

Balance this equation

MnO2 + Al → Al2O3 + Mn

D 200

3MnO2 + 4Al → 2Al2O3 + 3Mn

D 300

What is meant by a balanced equation

D 300

Obeying the law of conservation of matter: same

number of atoms of each element in reactants and

products

D 400

Write a complete chemical equation for

aluminum sulfate + sodium hydroxide → aluminum hydroxide + sodium sulfate

D 400

Al2(SO4)3 +6NaOH 2Al(OH)3 + 3Na2SO4

D 500

Write the combustion reaction for

C2H4ç

D 500

C2H4 + 3O2 2H2O + 2CO2

E 100

Reaction Type?

C10H8 + 12O2 10CO2 + 4H2O

E 100

Combustion

E 200

Reaction Type?

2H2O 2H2 + O2

E 200

Decomposition

E 300

Reaction Type?

Mg + 2H2O Mg(OH)2 + H2

E 300

Single Replacement

E 400

Reaction Type?

8 Fe + S8 8 FeS

E 400

Combination

E 500

Reaction Type?

Pb(NO3)2 + 2KI PbI2 + 2KNO3

E 500

Double Replacement

F 100

Al + SnCl2 →

F 100

2Al + 3SnCl2 → 3Sn + 2AlCl3

F 200

Cl2 + NaBr →

F 200

Cl2 + 2NaBr → 2NaCl + Br2

F 300

Pb(NO3)2 + K2CrO4 →

F 300

Pb(NO3)2 + K2CrO4 → PbCrO4 + 2KNO3

F 400

Cu + ZnSO4

F 400

No Reaction!

Copper is less reactive than zinc

F 500

A piece of magnesium is dropped into hydrochloric acid (aqueous hydrogen

chloride)

F 500

Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq)

Final Jeopardy

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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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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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