The document outlines topics covered in General Chemistry II for week 4, including: 1) Expressing concentration of solutions using various units 2) Performing stoichiometric calculations for reactions in solution 3) Describing the effect of concentration on colligative properties of solutions It then provides examples of stoichiometric calculations involving reactions in aqueous solution, determining the amounts of reactants and products based on the amounts given of one substance.

1. GENERAL CHEMISTRY II
WEEK 4

2. 1. Use different ways of expressing
concentration of solutions: percent by mass,
mole fraction, molarity, molality, percent by
volume, percent by mass, ppm.
2. Perform stoichiometric calculations for
reactions in solution.
3. Describe the effect of concentration on the
colligative properties of solutions .

3. 4. Differentiate colligative properties of
nonelectrolyte solutions and of electrolyte
solution.
5. Calculate the boiling point elevation and
freezing point depression from the
concentration of a solute in a solution

4. 6. Calculate molar mass from colligative
property data.
7. Describe laboratory procedures in
determining concentration of solution.

5. General Chemistry 2 – Senior High School (STEM)
EQ: Why is solution stoichiometry significant?

6.  The quantitative study of the relative
amounts of reactants and products in
chemical reactions is referred to
as stoichiometry.
 Solution stoichiometry deals with relative
quantities of reactants and products for
chemical reactions occurring in solutions.

7. 1. How many grams of solid calcium
hydroxide, Ca(OH)2, are required to react
with 350 mL of 0.40 M HCl?
__HCl + __Ca(OH)2  __CaCl2 + __ H2O

8. 1. How many grams of solid calcium hydroxide,
Ca(OH)2, are required to react with 350 mL of
0.40 M HCl?
__HCl + __Ca(OH)2  __CaCl2 + __ H2O
Step 1: Identify the given and unknown units.
Given: _________________________
Find: _________________________

9. 1. How many grams of solid calcium hydroxide,
Ca(OH)2, are required to react with 350 mL of
0.40 M HCl?
__HCl + __Ca(OH)2  __CaCl2 + __ H2O
Step1: Identify the given and unknown units.
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2

10. Step 2 : Balance the chemical equation:
__HCl + __Ca(OH)2  __CaCl2 + __ H2O

11. Step 2 : Balance the chemical equation:
2HCl + __Ca(OH)2  __CaCl2 + 2H2O

12. Step 3: Convert the given mL to L
Given: 350 mL of 0.40 M HCl

13. Step 3: Convert the given mL to L
Given: 350 mL of 0.40 M HCl
350 mL x
𝟏 𝑳
𝟏𝟎𝟎𝟎 𝒎𝑳
= 0.35o L

14. Step 4: Write the “given” and “unknown units”
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2
0.350L x x x = ____ g of Ca(OH)2

15. Step 5: Fill in factors
2HCl + __Ca(OH)2  __CaCl2 + 2H2O
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2
0.350L x
𝟎.𝟒𝟎 𝒎𝒐𝒍 𝑯𝑪𝒍
𝟏 𝑳
x x = ____ g of Ca(OH)2

16. Step 5: Fill in factors
2HCl + __Ca(OH)2  __CaCl2 + 2H2O
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2
0.350L x
𝟎.𝟒𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐋
x
𝟏 𝐦𝐨𝐥 Ca(OH)2
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
x = ____ g of
Ca(OH)2

17.  Ca(OH)2
Ca - 1 x 40 = 40g/mol
O - 2 x 16 = 32g/mol
H - 2 x 1 = 2 g/mol
Total Molar Mass = 74g/mol

18. Step 5: Fill in factors
2HCl + __Ca(OH)2  __CaCl2 + 2H2O
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2
0.350L x
𝟎.𝟒𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐋
x
𝟏 𝐦𝐨𝐥 Ca(OH)2
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
x
𝟕𝟒 𝐠
𝟏 𝐦𝐨𝐥
= ____ g of
Ca(OH)2

19. Step 6: Solve
2HCl + __Ca(OH)2  __CaCl2 + 2H2O
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2
0.350L x
𝟎.𝟒𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐋
x
𝟏 𝐦𝐨𝐥 Ca(OH)2
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
x
𝟕𝟒 𝐠
𝟏 𝐦𝐨𝐥
= ____ g of
Ca(OH)2

20. Step 6: Solve
2HCl + __Ca(OH)2  __CaCl2 + 2H2O
Given: 350 mL of 0.40 M HCl
Find: grams of Ca(OH)2
0.350L x
𝟎.𝟒𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐋
x
𝟏 𝐦𝐨𝐥 Ca(OH)2
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
x
𝟕𝟒 𝐠
𝟏 𝐦𝐨𝐥
= 5.18 g of
Ca(OH)2

21. 2. How many grams of aluminum are required
to react with 35 mL of 2.0 M hydrochloric
acid, HCl?
__ HCl + __ Al  __ AlCl3 + __ H2

22. 2. How many grams of aluminum are required
to react with 35 mL of 2.0 M hydrochloric
acid, HCl?
__ HCl + __ Al  __ AlCl3 + __ H2
Step 1: Identify the given and unknown units.
Given: _________________________
Find: _________________________

23. 2. How many grams of aluminum are required
to react with 35 mL of 2.0 M hydrochloric
acid, HCl?
__ HCl + __ Al  __ AlCl3 + __ H2
Step 1: Identify the given and unknown units.
Given: 35 mL of 2.0M HCl
Find: g of Al

24. Step 2 : Balance the chemical equation:
__ HCl + __ Al  __ AlCl3 + __ H2

25. Step 2 : Balance the chemical equation:
6 HCl + 2 Al  2 AlCl3 + 3H2

26. Step 3: Convert the given mL to L
Given: 35 mL of 2.0M HCl

27. Step 3: Convert the given mL to L
Given: 35 mL of 2.0M HCl
35 mL x
𝟏 𝑳
𝟏𝟎𝟎𝟎 𝒎𝑳
= 0.035 L

28. Step 4: Write the “given” and “unknown units”
Given: 35 mL of 2.0M HCl
Find: g of Al
0.035L x x x = ____ g of Al

29. Step 5 : Fill in factors
6 HCl + 2 Al  2 AlCl3 + 3H2
Given: 35 mL of 2.0M HCl
Find: g of Al
0.035L x
𝟐.𝟎 𝒎𝒐𝒍 𝑯𝑪𝒍
𝟏 𝑳
x x = ____ g of Al

30. Step 5 : Fill in factors:
6 HCl + 2 Al  2 AlCl3 + 3H2
Given: 35 mL of 2.0M HCl
Find: g of Al
0.035L x
𝟐.𝟎 𝒎𝒐𝒍 𝑯𝑪𝒍
𝟏 𝑳
x
𝟐 𝒎𝒐𝒍 𝑨𝒍
𝟔 𝒎𝒐𝒍 𝑯𝑪𝒍
x = ____ g of Al

31. Al = 1 x 27 = 27 g/mol

32. Step 5 : Fill in factors:
6 HCl + 2 Al  2 AlCl3 + 3H2
Given: 35 mL of 2.0M HCl
Find: g of Al
0.035L x
𝟐.𝟎 𝒎𝒐𝒍 𝑯𝑪𝒍
𝟏 𝑳
x
𝟐 𝒎𝒐𝒍 𝑨𝒍
𝟔 𝒎𝒐𝒍 𝑯𝑪𝒍
x
𝟐𝟕 𝒈
𝟏 𝒎𝒐𝒍
= ____ g of Al

33. Step 6 : Solve
6 HCl + 2 Al  2 AlCl3 + 3H2
Given: 35 mL of 2.0M HCl
Find: g of Al
0.035L x
𝟐.𝟎 𝒎𝒐𝒍 𝑯𝑪𝒍
𝟏 𝑳
x
𝟐 𝒎𝒐𝒍 𝑨𝒍
𝟔 𝒎𝒐𝒍 𝑯𝑪𝒍
x
𝟐𝟕 𝒈
𝟏 𝒎𝒐𝒍
= ____ g of Al

34. Step 6 : Solve
6 HCl + 2 Al  2 AlCl3 + 3H2
Given: 35 mL of 2.0M HCl
Find: g of Al
0.035L x
𝟐.𝟎 𝒎𝒐𝒍 𝑯𝑪𝒍
𝟏 𝑳
x
𝟐 𝒎𝒐𝒍 𝑨𝒍
𝟔 𝒎𝒐𝒍 𝑯𝑪𝒍
x
𝟐𝟕 𝒈
𝟏 𝒎𝒐𝒍
= 0.63 g of Al

35. 3. How many liters of a 3.0 M H3PO4 solution
are required to react with 4.5 g of zinc?
__ H3PO4 + __ Zn  __ Zn3(PO4)2 + __ H2

36. 3. How many liters of a 3.0 M H3PO4 solution
are required to react with 4.5 g of zinc?
__ H3PO4 + __ Zn  __ Zn3(PO4)2 + __ H2
Step 1: Identify the given and unknown units.
Given: __________
Find: __________

37. 3. How many liters of a 3.0 M H3PO4 solution
are required to react with 4.5 g of zinc?
__ H3PO4 + __ Zn  __ Zn3(PO4)2 + __ H2
Step 1: Identify the given and unknown units.
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4

38. Step 2: Balance the chemical equation.
__ H3PO4 + __ Zn  __ Zn3(PO4)2 + __ H2

39. Step 2: Balance the chemical equation.
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2

40. Step 3: Write the “given” and “unknown units”
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4
4.5 g Zn x x x = _____ L H3PO4

41. Step 4: Fill in factors
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4
4.5 g Zn x x x = _____ L H3PO4

42.  4.5 grams of Zn
Zn : 1 x 65 = 65 g/mol

43. Step 4: Fill in factors
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4
4.5 g Zn x
𝟏 𝒎𝒐𝒍
𝟔𝟓 𝒈 𝒁𝒏
x x = _____ L H3PO4

44. Step 4: Fill in factors
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4
4.5 g Zn x
𝟏 𝒎𝒐𝒍
𝟔𝟓 𝒈 𝒁𝒏
x
𝟐 𝒎𝒐𝒍 H3PO4
𝟑 𝒎𝒐𝒍 𝒁𝒏
x
𝟏 𝑳
𝟑.𝟎 𝒎𝒐𝒍H3PO4
= _____ L H3PO4

45. Step 5: Solve
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4
4.5 g Zn x
𝟏 𝒎𝒐𝒍
𝟔𝟓 𝒈 𝒁𝒏
x
𝟐 𝒎𝒐𝒍 H3PO4
𝟑 𝒎𝒐𝒍 𝒁𝒏
x
𝟏 𝑳
𝟑.𝟎 𝒎𝒐𝒍H3PO4
= _____ L H3PO4

46. Step 5: Solve
2 H3PO4 + 3 Zn  Zn3(PO4)2 + 3H2
Given: 4.5 g of Zn
Find: L of a 3.0 M H3PO4
4.5 g Zn x
𝟏 𝒎𝒐𝒍
𝟔𝟓 𝒈 𝒁𝒏
x
𝟐 𝒎𝒐𝒍 H3PO4
𝟑 𝒎𝒐𝒍 𝒁𝒏
x
𝟏 𝑳
𝟑.𝟎 𝒎𝒐𝒍H3PO4
= 0.015 L H3PO4

47. 4. How many milliliters of 1.50 M Nitric acid is
required to react with 100.0 g of cuprous
oxide?
HNO3 + Cu2O  Cu(NO3)2 + NO + H2O

48. 4. How many milliliters of 1.50 M Nitric acid is
required to react with 100.0 g of cuprous oxide?
HNO3 + Cu2O  Cu(NO3)2 + NO + H2O
Step 1: Identify the given and unknown units.
Given: __________
Find: __________

49. 4. How many milliliters of 1.50 M Nitric acid
(HNO3) is required to react with 100.0 g of
cuprous oxide (Cu2O)?
HNO3 + Cu2O  Cu(NO3)2 + NO + H2O
Step 1: Identify the given and unknown units.
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3

50. Step 2: Balance the chemical equation.
HNO3 + Cu2O  Cu(NO3)2 + NO + H2O

51. Step 2: Balance the chemical equation.
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O

52. Step 3: Write the “given” and “unknown units”
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3
100.0 g Cu2O x x x = _____ mL
HNO3

53. Step 4: fill in factors
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3
100.0 g Cu2O x x x = _____ mL
HNO3

54.  100.0 g of Cu2O
Cu : 2 x 64 g = 128 g/mol
O : 1 x 16 g = 16 g/mol
Total Molar Mass = 144 g/mol

55. Step 4: fill in factors
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3
100.0 g Cu2O x
𝟏 𝒎𝒐𝒍
𝟏𝟒𝟒 𝒈
x x = _____ L
HNO3

56. Step 4: fill in factors
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3
100.0 g Cu2O x
𝟏 𝒎𝒐𝒍
𝟏𝟒𝟒 𝒈
x
𝟏𝟒 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑
𝟑 𝒎𝒐𝒍 𝑪𝒖𝟐
𝑶
x
= ___L HNO3

57. Step 4: fill in factors
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3
100.0 g Cu2O x
𝟏 𝒎𝒐𝒍
𝟏𝟒𝟒 𝒈
x
𝟏𝟒 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑
𝟑 𝒎𝒐𝒍 𝑪𝒖𝟐
𝑶
x
𝟏 𝑳
𝟏.𝟓 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑
= ___L HNO3

58. Step 5: Solve
14HNO3 + 3Cu2O  6Cu(NO3)2 + 2NO + 7H2O
Given: 100.0 g of Cu2O
Find: mL of 1.50 M HNO3
100.0 g Cu2O x
𝟏 𝒎𝒐𝒍
𝟏𝟒𝟒 𝒈
x
𝟏𝟒 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑
𝟑 𝒎𝒐𝒍 𝑪𝒖𝟐
𝑶
x
𝟏 𝑳
𝟏.𝟓 𝒎𝒐𝒍 𝑯𝑵𝑶𝟑
= 2.16 L HNO3

59. Step 6: Convert L to mL
2.16 L x
𝟏𝟎𝟎𝟎 𝒎𝑳
𝟏𝑳
= 2160 mL HNO3