1. Stoichiometry
Additional
Examples

2. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?

3. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
2 H2 + O > 2 H2O

4. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
2 H2 + O > 2 H2O
(3.3 mol O2)
#g H2 =

5. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
2 H2 + O > 2 H2O
(3.3 mol O2)
#g H2 = (1 mol O2)

6. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
2 H2 + O > 2 H2O
(3.3 mol O2) (2 mol H2)
#g H2 = (1 mol O2)
stoichiometric
factor

7. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
2 H2 + O > 2 H2O
(3.3 mol O2) (2 mol H2) (2.0 g H2)
#g H2 = (1 mol O2) (1 mol H2)
the molar mass of H2

8. EXAMPLE
What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
2 H2 + O > 2 H2O
(3.3 mol O2) (2 mol H2) (2.0 g H2)
#g H2 = (1 mol O2) (1 mol H2)
= 13 g H2

9. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?

10. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O

11. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O
(5.40 g H2O)
#g H2 =

12. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O
(5.40 g H2O) (1 mol H2O)
#g H2 = (18.0 g H2O)
molar mass of H2O

13. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O
(5.40 g H2O) (1 mol H2O) (2 mol H2)
#g H2 = (18.0 g H2O)

14. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O
(5.40 g H2O) (1 mol H2O) (2 mol H2)
#g H2 = (18.0 g H2O) (2 mol H2O)
stoichiometric
factor

15. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O
(5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2)
#g H2 =
(18.0 g H2O) (2 mol H2O) (1 mol H2)
molar mass H2

16. EXAMPLE
What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
2 H2 + O > 2 H2O
(5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2)
#g H2 =
(18.0 g H2O) (2 mol H2O) (1 mol H2)
= g H2

17. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.

18. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?

19. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
KClO > O2 + ?

20. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
KClO > O2 + ?
? => KCl
KClO > O2 + KCl

21. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
potassium chlorate -----> oxygen + ?
KClO > O2 + ?
? => KCl
KClO > O2 + KCl
2 KClO > 3 O KCl

22. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO > 3 O KCl
(100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 =
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)

23. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO > 3 O KCl
(100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 =
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass

24. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO > 3 O KCl
(100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 =
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass
stoichiometric
factor

25. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO > 3 O KCl
(100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 =
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass molar mass
stoichiometric
factor

26. EXAMPLE
Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
2 KClO > 3 O KCl
(100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2)
#g O2 =
(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
molar mass molar mass
= 39.2 g O2
stoichiometric
factor

27. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?

28. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
nitrogen + hydrogen -----> ammonia

29. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
nitrogen + hydrogen -----> ammonia
N2 + H > NH3

30. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
nitrogen + hydrogen -----> ammonia
N2 + H > NH3
N H > 2 NH3

31. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N H > 2 NH3
(3.0 kg NH3)
#mol H2 =

32. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N H > 2 NH3
(3.0 kg NH3) (1000 g)
#mol H2 =
(1 kg)

33. EXAMPLE
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N H > 2 NH3
(3.0 kg NH3) (1000 g) (1 mol NH3)
#mol H2 =
(1 kg) (17.0 g NH3)

34. EXAMPLE (3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2)
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N H > 2 NH3
(3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2)
#mol H2 =
(1 kg) (17.0 g NH3) (2 mol NH3)
stoichiometric
factor

35. EXAMPLE (3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2)
Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
N H > 2 NH3
(3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2)
#mol H2 =
(1 kg) (17.0 g NH3) (2 mol NH3)
= 2.6 X 102 mol H2

36. EXAMPLE
The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?

37. EXAMPLE
The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide

38. EXAMPLE
The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
ZnS + O > ZnO + SO2

39. EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
ZnS + O > ZnO + SO2
2 ZnS + 3 O > 2 ZnO + 2 SO2

40. EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O > 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2)
#g O2 =
(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
stoichiometric
factor

41. EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O > 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2)
#g O2 =
(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
= 3.45 g O2

42. EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O > 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2)
#gSO2 =
(97.44 g ZnS) (2 mol ZnS) (1 mol SO2)

43. EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
2 ZnS + 3 O > 2 ZnO + 2 SO2
(7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2)
#gSO2 =
(97.44 g ZnS) (2 mol ZnS) (1 mol SO2)
= 4.60 g SO2

44. EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide.
Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
#g O2 = 3.45 g O2
#gSO2 = g SO2

45. EXAMPLE
Zinc and sulfur react to form zinc sulfide, a substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is Zn + S > ZnS In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.

46. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) Which is the limiting reactant? (b) How many grams of ZnS can be formed from this particular reaction mixture? (c) How many grams of which element will remain unreacted in this experiment?

47. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn
(12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS)
#g ZnS =
(65.38 g Zn) (1 mol Zn) (1 mol ZnS)

48. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn
(12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS)
#g ZnS =
(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
= 17.9 g ZnS

49. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S
(6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS)
#g ZnS =
(32.06 gS ) (1 mol S) (1 mol ZnS)

50. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S
(6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS)
#g ZnS =
(32.06 gS ) (1 mol S) (1 mol ZnS)
= 19.8 g ZnS

51. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.9 g ZnS can be produced.

52. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.9 g ZnS can be produced.
(c) How many grams of which element will remain unreacted in this experiment?
if use all 12.0 g Zn

53. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn
(12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S)
#g S =
(65.38 g Zn) (1 mol Zn) (1 mol S)

54. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn
(12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S)
#g S =
(65.38 g Zn) (1 mol Zn) (1 mol S)
= 5.88 g S reacted

55. EXAMPLE In a particular experiment 12. 0 g of Zn are mixed with 6
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S > ZnS
(a) The limiting reactant? (b) #grams of ZnS?
if use all 12.0 g Zn #g ZnS = g ZnS
if use all 6.50 g S #g ZnS = 19.8 g ZnS
Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn #g S = 5.88 g S reacted
Therefore, ( )g = 0.61 g S remain unreacted.

56. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?

57. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

58. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= ( *35.453)amu

59. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= ( *35.453)amu
= amu

60. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= ( *35.453)amu
= amu
1(gaw)
%C = X 100 MM

61. MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= ( *35.453)amu
= amu
1(gaw)
%C = X 100 MM
1(12.011)
%C = X 100 = % C

62. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= ( *35.453)amu
= amu
1( )
%H = X 100 = % H

63. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= ( *35.453)amu
= amu
3(35.453)
%Cl = X 100 = % Cl

64. EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= amu
%C = % C
%H = % H
%Cl = % Cl