Chapter 11 Review

Chapter 10 Review

Magnesium Oxide Lab

Page #727-730 Problems #12-15:

12. HBr
13. 2HI ---> H2+I2
14. 3Mg+N2 ---> Mg3N2
15. a) Fe+Pb(NO3)2 ---> Fe(NO3)2+Pb
      b) Cl2+2NaI ---> 2NaCl+I2
      c) Ca+2H2O ---> Ca(OH)2+H2
      d) Zn+H2SO4 ---> ZnSO4+H2

Page #733 Problems #16-17:

16. a) 3NaOH+Fe(NO3)3 ---> Fe(OH)3+3NaNO3
      b) 3Ba(NO3)2+2H3PO4 ---> Ba3(PO4)2+6HNO3
      c) FeS+2HCl ---> H2S+FCl2
17. a) 3KOH+H3PO4 ---> K3PO4+3H2O
      b) AgNO3+NaCl ---> AgCl+NaNO3
      c) 3Ca(OH)2+2H3PO4 ---> Ca(PO4)2+6H2O
      d) 2KI+Pb(NO3)2 ---> 2KNO3+PbI2
      e) 3H2SO4+2Al(OH)3 ---> Al2(SO4)3+6H2O

Types of Chemical Reactions Pogil

Percent Composition of Hydrates Lab

Purpose:   to measure the percent of water in a series of crystalline compounds called hydrates.

Procedure: 

1. Label each tube with and measure/record the masses

2. Add corresponding compound (2-3g) measure and record mass of each tube and compound

3. Heat contents of test tube and note any changes in appearance of the solid contents

4. Continue heating until moisture is driven from the tube. Heat for 2-3 minutes. Repeat for the other test tubes

5. Allow each test tube to cool, then measure and the mass of each test tube and the heated compound.

Data: 

Tube 1: Calcium dihydrate-
1. Before heat: 2.5 g After heat: 7.45 - 7.39= 0.06 g
2. Difference: 2.44g
3. % water loss: .06/2.5x100= 2%

Tube 2: Sodium sulfate decahydrate-
1. Before heat: 2.6 g After heat: 7.93 - 7.39= .54g
2. Difference: 2.6-0.54= 2.06g
3. % water lost: 0.54/2.6x100= 21%

Tube 3: Copper II sulfate pentahydrate-
1. Before heat: 3.1 g After heat: 7.5 g
2. Difference: 3.1-0.11= 2.99g
3. % water loss: .11/3.1x100= 4%

Observations

Tube 1: no change 

Tube 2: quickly melting, fizzing, changing to liquid, bubbling up, separates when taken off flame 

Tube 3: turning white, flame turns green, turning grey because it is burning, has a burnt smell

4. Sodium sulfate decahydrate lost the

Check:

H2O lost:

Tube 1: 2g(1 mol/18g)= .112 mol 

Tube 2: 21g(1 mol/18g)= 1.167 mol

Tube 3: 4g(1 mol/18g)= .222 mol

Compounds: 

Tube 1: 98g(1mol/110g) = .89 mol

Tube 2: 79g(1mol/142g) = .56 mol

Tube 3: 96g(1mol/250g) = .38 mol

CuSO45H2O

Na2SO410H2O

Na2CO32H2O

Balancing Equations

Pages #672-673 Problems #41-42:

41. C3Cl3N3
42. C2H6O2

Page #662 Problems #33-40:

33. 72.18% Mg, 27.82% N
34. 93% Hg, 7% O
35. a) 82.35% N
      b) 35% N
36. a) 81% C, 19% H
      b) 19% Na 1% H 27% S 53% O
37. a) 102.5g N
      b) 43.75g N
38. a) 66.5g H
      b) 3.84g H
39. a)OH
      b)HgSO4
40. C3H8N

Page #656 Problems #16-23:

16.  1.27 g

17. 182.5 g
18. .034 mol
19. .099 mol
20. a) .072 LCO2
      b) 82.88 LN2
      c) 21.504 LCH4
21. a) 3LSO2
      b) .039LHe
      c) 44.643 LC2H6
22. 80.192 g/mol
23. 3.75 g/L