Pg. 727 #12-14

12. HBr

13. HI --> H+I

14. 3Mg+N2 --> Mg3N2

Pg. 733 #16 & 17

16.

a) 3NaNO3 + Fe(OH)3

b) Ba3(PO4)2 + 6HNO3

c) FeS + HCl

17.

a) KOH + H3PO4 --> KPO4 + H2OH

b) AgNO3 + NaCl --> AgCl + NaNO3

c) Ca(OH)2 + H3PO4 --> CaPO4 + H2OH

d) Kl +Pb(NO3)2 --> KNO3+Pb2

e) H2SO4+Al(OH)3 --> H2OH+ASO4

Lab

Purpose: To measure the % of water in a series of crystalline compounds called hydrates.

Procedure: 

1. label each tube with and record the masses

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

3. Heat contents of test tube and record 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- 2.5g

Tube 2: Sodium sulfate decahydrate- 2.6g

Tube 3: copper II sulfate pentahydrate- 3.1g

Observations:

Test 1- calcium dihydrate- not affected by heat- 7.45g

Test 2- sodium sulfate- dissolving, fizzing, melting, boiling, separates when departed from heat- 7.93g

Test 3- copper II sulfate pentahydrate- Changes to white, burns, omits green flame, as a result of burning, color changes to grey- 7.50g

Empty tube: 7.39

Calcium dihydrate: 

--Before- 2.5g

After- 7.45-7.39= 0.06g

--Difference- 2.5-0.06= 2.44g 

--% water lost- .06/2.5X100= 2%

Sodium sulfate: 

--Before- 2.6g

After- 7.93-7.39= 0.54g

--Difference- 2.6-0.54= 2.06g

--% water lost- 0.54/2.6 X100= 21%

Copper II sulfate pentahydrate:

--Before- 3.1g

After- 7.50-7.39= 0.11g

--Difference (water lost) - 3.1-0.11g= 2.99g

--% water lost- .11/3.1X100= 4%

Sodium sulfate lost the greatest amount of water in grams at 21%

Calcium dihydrate lost the least amount of water in  grams at 2% 

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

Balancing equations

Pg. 672 #41&42

41. 3CHO

42. 2CH3O

Pg. 662-671 #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 B.) 43.75g N 

38. A.) 66.5g H B.) 3.84g H

39. A.) OH B.) HgSO4

40. 2CHN3

Pg. 648-656 #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