1. By Carl Lyman © September 2001
Jeopardy Template
By Carl Lyman
© September 2001

2. Midterm Jeopardy 2 100 200 300 400 500 MOF Reactions Stoichiometry
Thermo
100
200
300
400
500

3. Credits
That’s all folks.
Exit

4. MOF $100 What is the molar mass of (NH4)2CO3? A.) 144 g/mol
B.) 138 g/mol
C.) 96 g/mol
D.) 78 g/mol

5. MOF $100 C 96 g/mol Subscript must be multiplied in for parentheses
N 2 x 14 = 28
H 8 x 1 = 8
C 1 x 12 = 12
O 3 x 16 = 48

6. MOF $200 What is the percent composition of chromium in BaCrO4?
D.) 25.2 %

7. MOF $200 C 20.5% Ba 1 x 137 = 137 Part/Whole x 100 Cr 1 x 52 = 52
O 4 x 16 = x 100 =
253 g/mol

8. MOF $300 What is the mass in grams of 5.90 moles of C8H18?
A.) g
B.) 19.4 g
C.) 389 g
D.) 673 g

9. MOF $300 D 673 g 5.90 mol x 114 g = 1 mol C 8 x 12 = 96 H 18 x 1 = 18

10. How many moles of silver atoms are in 1.8 x 1020 atoms of silver?
MOF $400
How many moles of silver atoms are in 1.8 x 1020 atoms of silver?
A.) 3.0 x 10-4
B.) 3.3 x 10-3
C.) 3.0 x 102
D.) 1.1 x 1044

11. MOF $400 A 3.0 x 10-4 mol 1.8 x 1020 atoms x 1 mol = 6.02 x 1023 atoms
Don’t forget to use parentheses when plugging scientific notation into your calculator

12. MOF $500 Which of the following is NOT an empirical formula? A.) Sb2S3
B.) C2H8O
C.) C2N2H8
D.) BeCr2O7

13. MOF $500
C C2N2H8
Empirical formulas are in lowest whole number ratio of atoms. C2N2H8 can be reduced to CNH4

14. CaSO4 + AlCl3  Al2(SO4)3 + CaCl2
Reactions $100
CaSO4 + AlCl3  Al2(SO4)3 + CaCl2
What is the coefficient of Al2(SO4)3 when the equation is completely balanced using the smallest whole-number coefficients?
A.) 1
B.) 2
C.) 3
D.) 4

15. Reactions $100 A 1 3CaSO4 + 2AlCl3  Al2(SO4)3 + 3CaCl2
Balanced equations have the same number of atoms of each element in both the reactants and products. Only coefficients can be altered.

16. is an example of which type of reaction?
Reactions $200
The equation
2C3H7 + 9O2  6CO2 + 8H2O
is an example of which type of reaction?
A.) single-replacement reaction
B.) double-replacement reaction
C.) decomposition reaction
D.) combustion reaction

17. Reactions $200 D combustion reaction
A hydrocarbon reacts with oxygen to produce carbon dioxide and water

18. is an example of which type of reaction?
Reactions $300
The equation
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
is an example of which type of reaction?
A.) combination reaction
B.) single-replacement reaction
C.) decomposition reaction
D.) double-replacement reaction

19. Reactions $300 B single replacement reaction
The free element Mg replaces hydrogen in the compound HCl.

20. Reactions $400
A double-replacement reaction takes place when aqueous Na2CO3 reacts with aqueous Sn(NO3)2. You would expect one of the products of this reaction to be ________.
A.) NaNO3
B.) NaSn
C.) Sn(CO3)2
D.) CNO3

21. Reactions $400 A NaNO3 Na2CO3 + Sn(NO3)2  SnCO3 + 2NaNO3
Na and Sn trade places

22. Reactions $500 Predict the products: Mg + CuCl2  A.) No reaction
B.) MgCu + Cl2
C.) Cu + MgCl2
D.) MgCuCl2

23. Reactions $500
C Cu + MgCl2
This is a single replacement reaction. It will occur because the free element Mg is higher in activity than the element it is replacing in the compound. Mg and Cu will be the ones trading places because they are both metals and have positive oxidation numbers.

24. Stoichiometry $100 6 CO2 + 6 H2O  C6H12O6 + 6 O2
What is the total number of moles of water needed to make 2.5 moles of C6H12O6?
A.) 2.5
B.) 6.0
C.) 12
D.) 15

25. Stoichiometry $100
D 15 mol
2.5 = X
← From original equation

26. 2Pb(NO3)2(s)  2PbO(s) + 4NO2(g) + O2(g)
Stoichiometry $200
2Pb(NO3)2(s)  2PbO(s) + 4NO2(g) + O2(g)
How many grams of oxygen are produced when 11.5 grams of NO2 is formed?
A.) 1.00 g
B.) 2.00 g
C.) 2.88 g
D.) 32.0 g

27. Stoichiometry $200 B 2.00 g 11.5g x 1mol 0.25 = X 0.0625mol x 32g =
46g mol

28. Stoichiometry $300 2 C4H10 + 13 O2  8 CO2 + 10 H2O
What is the total number of moles of O2 that must react completely with 5.00 moles of C4H10?
A.) 10.0
B.) 20.0
C.) 26.5
D.) 32.5

29. Stoichiometry $300
D mol
5.00 = X

30. Stoichiometry $400 4Fe(s) + 3O2(g) 2Fe2O3(s)
How many grams of Fe2O3 are formed when 16.7 grams of Fe reacts completely with oxygen?
A.) 12.0 g
B.) 23.9 g
C.) 47.8 g
D.) 95.6 g

31. Stoichiometry $400 B 23.9 g 16.7g x 1mol 0.298 = X 0.149mol x 160g =
56g mol

32. Stoichiometry $500 Ca + F2  CaF2
What is the limiting reagent and what quantity of CaF2 results from the reaction of 3.00 g of calcium and 2.00 g of fluorine?
A.) Ca, 4.78
B.) F2, 4.78
C.) Ca, 4.11
D.) F2, 4.11
E.) Ca, 3.22

33. Stoichiometry $500
D F2, 4.11
Determine how many grams of product can be made
Ca 3g x 1mol = X mol x 78g = 5.85g
40g mol
F2 2g x 1mol = X mol x 78g = 4.11g
38g mol
F2 is the limiting reagent because it can make the least amount of product. The theoretical yield is the smaller amount of product made and is based on the amount made by the limiting reagent.

34. Which Kelvin temperature is equal to -73°C?
Thermo $100
Which Kelvin temperature is equal to -73°C?
A.) 100 K
B.) 173 K
C.) 200 K
D.) 346 K

35. Thermo $100
C 200 K
K = °C + 273 =

36. Thermo $200
The average kinetic energy of water molecules is greatest in which of these samples?
A.) 10 g of water at 35°C
B.) 10 g of water at 55°C
C.) 100 g of water at 25°C
D.) 100 g of water at 45°C

37. Thermo $200 B 10g of water at 55°C
Temperature measures kinetic energy, the higher the temperature the higher the kinetic energy

38. Thermo $300
The specific heat of silver is 0.24 J/gºC. How much heat is needed to warm 4.37 g of silver from 25.0ºC to 27.5ºC?
A.) 2.62 J
B.) 0.14 J
C.) 45.5 J
D.) J

39. Thermo $300
A J
q = mcΔT
q = (4.37 g)(0.24 J/g°C)(27.5°C – 25.0°C)

40. Calculate the energy required to produce 7 moles of Cl2O7
Thermo $400
2Cl2 + 7 O kJ  2 Cl2O7
Calculate the energy required to produce 7 moles of Cl2O7
A.) kJ
B.) kJ
C.) kJ
D.) kJ

41. Thermo $400
D kJ
7 = X
← From original equation

42. Thermo $500
Given the potential energy diagram of a chemical reaction: (see image)
Which arrow represents the potential energy of the reactants?
A.)
B.)
C.)
D.)

43. Category 4 $500 B A = activation energy C = P.E. of activated complex
D = P.E. of products