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Chapter 12 CHEMICAL STOICHIOMETRY

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1 Chapter 12 CHEMICAL STOICHIOMETRY
Reactants: Zn + I2 Product: Zn I2

2 Chapter 12: Stoichiometry I can ….. Statements
I can write and balance equations. I can calculate mass of products from mass of reactants and vice versa I can determine the limiting reactant in a chemical reaction.

3 STOICHIOMETRY is the calculation of relative quantities of reactants and products in chemical reactions.

4 Stoichiometry: The mathematical relationship between the reactants and the products of a chemical reactions. Stoichiometry uses mole ratios to calculate the amount in grams of: Product formed Or reactant needed

5 STOICHIOMETRY 2 Al(s) + 3 Br2(l)  Al2Br6(s)
It rests on the principle of conservation of matter. 2 Al(s) + 3 Br2(l)  Al2Br6(s)

6 Law of Conservation of Mass
Matter cannot be created or destroyed, so the mass of the reactants must equal the mass of the products in a chemical reaction

7 2 CO + O2  2 CO2 88g reactants = 88g products 1mol 2mol
1 (32g) 2 (44g) 32g 88g 2mol 2 (28g) 56g 88g reactants = 88g products

8 O2  2 CO2 2 CO + The coefficients represent the number of moles of each substance in the equation. Balanced chemical equations show mole ratios among the products and reactants. Mass of reactants = mass of products (no mass is lost or gained!)

9 GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
grams reactant grams product Moles reactant Mole Ratio Moles product First balance the chemical equation!

10 Mole ratios Now lets make water molecules
Mole ratios Now lets make water molecules. The equation looks like this: The coefficients in the balanced equation represent the number of moles of each substance! So according to this equation , we need ____ moles of H2 and _____ moles of O2 to make ______ moles of H2O 2 2 1

11 Mole ratios Now lets make water molecules
Mole ratios Now lets make water molecules. The equation looks like this: What is the ratio of oxygen to water? (write as a fraction) If we only wanted to make one mole of water: How many moles of H2 would we need? How many moles of O2 would we need? 1 mol O2 2 mole H20 2 mole H20 1 mol O2

12 3) If we have 4 moles of H2 and 4 moles of O2, how many moles of water can we make?
4 mol H2 1 2 mol H2O = 4 mol H2O mol H2 4 mol O mol H2O mol O2 = 8 mol H2O Only 4 moles of water can be made because H2 runs out before O2 causing the reaction to stop.

13 4) If we have 4 moles of H2 and 2 moles of O2, how many moles of water can we make?
4 mol H2 1 2 mol H2O 2 mol H2 2 mol O mol H2O mol O2 Neither reactant runs out.

14 5. If we have 2 moles of H2 and 6 moles of O2, how many moles of water can we make?
2 mol H2 1 2 mol H2O = 2 mol H2O 2 mol H2 6 mol O2 1 2 mol H2O = 12 mol H2O 1 mol O2

15 Theoretical Yield: The maximum amount of product that can be formed by a chemical reaction. Percent Yield: The actual yield of a product expressed as a percentage of the theoretical yield

16 Actual yield x 100 = percent yield Theoretical yield
If a student's actually produces only 10.00 grams of product when the theoretical amount of product was calculated as grams, what is the student's % yield? 10.00g / g x 100 = 88.89% yield

17 Percent Yield Compares the experimental amount to the theoretical amount. It is a measure of the efficiency of your chemical reaction. Did the chemical reaction produce the expected amount of product? If not, why not? How can you modify the chemical reaction to improve the percent yield?

18 Steps to Solving Stoichiometry Problems
Step 1 – Write balanced chemical equation    STEP 2 - Convert the grams to moles. STEP 3 - Convert moles of reactant to moles of product STEP 4 - Convert moles back to grams.

19 Write the balanced equation NH4NO3  N2O + 2 H2O
PROBLEM 1: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products? PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products? STEP 1 Write the balanced equation NH4NO3  N2O + 2 H2O Chemical equations show mole ratios so all the masses of all substances must be converted to moles!

20 STEP 2 Convert the grams of reactant to moles of reactant
454 g of NH4NO3  N2O + 2 H2O 454 g of NH4NO3 � N2O + 2 H2O STEP 2 Convert the grams of reactant to moles of reactant 454 g 1 1 mol = 5.68 mol NH4NO3 80.04 g

21 STEP 3 Convert moles of reactant to moles of product
5.68 mol NH4NO3 2 mol H2O 1 1 mol NH4NO3 11.4 mol H2O produced

22 STEP 4 Convert moles of product to grams of product
11.4 mol H2O 18.02 g H2O = 204 g H2O 1 mol H2O 1 This is the THEORETICAL YIELD

23 STEP 5 Calculate the percent yield
If you isolated only 131 g of N2O, what is the percent yield? This compares the theoretical yield (250. g) and the actual yield (131 g) .

24 STEP 5 Calculate the percent yield
actual yield % yield = • 100% theoretical yield 131 g % yield = • 100% = 52.4% 250. g

25 Problem 2 Calcium metal reacts with water to form hydrogen gas. How many grams of Ca metal must be reacted to make 100 grams of hydrogen gas?

26 Calcium metal reacts with water to form hydrogen gas.
How many grams of Ca metal must be reacted to make 100 grams of hydrogen gas? Ca + 2 H2O  H2 + Ca(OH)2

27 Step 1 – Write balanced chemical equation
 We can use dimensional analysis to do steps 2 through 4. STEP 2 - Convert the mass to moles. STEP 3 - Convert moles of reactant to moles of product STEP 4 - Convert moles back to mass.

28 Ca + 2 H2O  H2 + Ca(OH)2 ? g 100 g

29 Ca + 2 H2O  H2 + Ca(OH)2 ? g 100 g 100 g H2 1

30 Ca + 2 H2O  H2 + Ca(OH)2 ? g 100 g 100 g H2 1 mol H2 1 2 g H2

31 Ca + 2 H2O  H2 + Ca(OH)2 ? g 100 g 100 g H2 1 mol Ca 1 mol H2

32 Ca + 2 H2O  H2 + Ca(OH)2 ? g 100 g 100 g H2 1 mol Ca 40.08 g Ca
1 mol H2 1 2 g H2 1 mol H2 1 mol Ca

33 Ca + 2 H2O  H2 + Ca(OH)2 ? g 100 g 100 g H2 1 mol Ca 40.08 g Ca 1 mol H2 1 2 g H2 1 mol H2 1 mol Ca 2004 grams of Ca is needed to make 100 grams of H2 gas !

34 Reactions Involving a LIMITING REACTANT
•• In a given reaction, there is not enough of one reagent to use up the other reagent completely. •• The reagent in short supply LIMITS the quantity of product that can be formed..

35 Limiting Reactant

36 The. amount of product you form depends. on. the amount of each
The amount of product you form depends on the amount of each reactant you have, and the mole ratio between reactants and products.

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43 The limiting reactant limits. the amount. of product. that
The limiting reactant limits the amount of product that can be formed in the reaction.

44 The limiting reactant is the substance that is completely consumed in the reaction. It controls the amount of product that can be produced. The excess reactant is the substance that is not completely consumed in the reaction and remains in excess after the reaction is finished.

45 N2 + 3H2  2 NH3 ? How much ammonia gas can be produce
from the reaction between 5.00 grams of nitrogen gas and 5.00 grams of hydrogen gas? N2 + 3H2  2 NH3 5.00 g ?

46 N2 + 3H2  2 NH3 5.00 g 5.00 g ?

47 N2 + 3H2  2 NH3 5.00 g 5.00 g ? 5.00g N2 1mol N2 2mol NH3 17.04g NH3 1mol N2 1mol NH3 28.02 g = 6.08 g NH3

48 N2 + 3H2  2 NH3 5 g 5 g ? 5.00g N2 1mol N2 2mol NH3 17.03g NH3
1mol H2 5.00g H2 2mol NH3 17.03 g NH3 1mol NH3 2.02g H2 3mol H2 = g NH3

49 6.08 g NH3 28.10 g NH3 Which of these is the correct amount of NH3 produced ? Why? Which substance is the limiting reactant in this reaction?

50 N2 + 3H2  2 NH3 5.00 g 5.00 g ? 5g N2 1mol N2 2mol NH3 17g NH3 1mol N2 1mol NH3 28 g = 6.08 g NH3 1mol H2 5g H2 2mol NH3 17g NH3 1mol NH3 2g H2 3mol H2 = g NH3

51 Which reactant is in excess and how much excess reactant remains unreacted at the end?
N2 + 3H2 5.00 g 5.00 g  2 NH3 6.08 g

52 By conservation of mass we see that only
Which reactant is in excess and how much excess reactant remains unreacted at the end? N2 + 3H2 5.00 g 5.00 g  2 NH3 6.08 g By conservation of mass we see that only 1.08 grams of H2 actually reacts

53 Which reactant is in excess and how much excess reactant remains unreacted at the end?
N2 + 3H2 5 g 5 g  2 NH3 6.08 g By conservation of mass we see that only 1.08 grams of H2 actually reacts So 5.00 g initial H2 – 1.08 g H2 reacted = g excess H2 is left unreacted

54 1mol N2 3mol H2 2.02g H2 1mol N2 1mol H2 28.02g = 1.08 grams H2 reacts
5.00g N2 1mol N2 3mol H2 2.02g H2 1mol N2 1mol H2 28.02g = 1.08 grams H2 reacts You can see that using the limiting reactant of grams of N2 , the stoichiometric calculation shows that 1.08 grams of H2 will react.

55 1.08 grams H2 reacts

56 1.07 grams H2 reacts 5.00g – 1.08g = 3.92g H2 remains unreacted

57 N2 + 3H2 5.00g 1.08g  2 NH3 6.08g 1.08 grams H2 reacts
We are left with 3.92 grams of excess hydrogen gas N2 + 3H g 1.08g  2 NH3 6.08g

58 N2 + 5.00g 3H2  2 NH3 5.00g 0g Before During 1.08g 6.08g 3H2  2 NH3 After 0g 3.92g 6.08g

59 2 H2 + O2  2 H2O 3.0g 10.0g ? Calculate the theoretical yield, determine the limiting reactant and excess reactant.

60 2 H2 + O2  2 H2O 3.0g 10.0g ? 3.0g H2 1mol H2 2mol H2O 2.02g H2
18.02g H2O 2.02g H2 2mol H2 1mol H2O = 26.8 g H2O

61 2 H2 + O2  2 H2O 3.0g 10.0g ? 3.0g H2 1mol H2 2g H2 2mol H2O 2mol H2
18g H2O 1mol H2O = 27.0 g H2O 10.0g O2 1mol O2 2mol H2O 18.02g H2O 32.00 g O2 1mol H2O = g H2O

62 2 H2 + O2  2 H2O 3.0g 10.0g ? 3.0g H2 1mol H2 2g H2 2mol H2O 2mol H2
18g H2O 1mol H2O = 27.0 g H2O 10g O2 1mol O2 2mol H2O 18g H2O 32 g O2 1mol H2O = g H2O

63 2 H2 + O2  2 H2O 3.0g 10.0g 11.26g 11.26g of H2O can be produced O2 is the limiting reactant 1.26 g H2 reacts 3.0g – 1.26g = 1.74 g excess H2 gas

64 11.26g of H2O can be produced O2 is the limiting reactant
= 1.26 g H2 reacts (11.26g – 10.0g = 1.26g) 3.0g – 1.26g = 1.74 g Excess H2 gas

65 2 H2 + O2  2 H2O 3.0g Before 10.0g 0g 2 H2 + O2  2 H2O 1.26g 10.00g 11.26g During 2 H2 + O2 1.74g 0g  2 H2O 11.26g After

66 31 grams of hydrogen gas reacts with 2. 89 L of oxygen gas
0.31 grams of hydrogen gas reacts with 2.89 L of oxygen gas. How much water is theoretically produced in the reaction? The density of oxygen gas is 1.31 g/L.

67 2 H2 + O2  2 H2O 0.31 g 2.89 L ?g

68 2 H2 + O2  0.31 g 2.89 L 2 H2O ?g 0.31 gH2 1mol 2mol H2O 2mol H2
18.02 g H2O 2.02 g H2 2mol H2 = 2.77 g H2O 2.89 L O2 1.31 g O2 1mol O2 2mol H2O 18.02 g 1L O2 32.00g O2 1mol H2O = 4.26 g H2O

69 If we only produce 1.38 grams of water in the reaction, what is the
% yield of the reaction?

70 If we only produce 1.38 grams of water in the reaction, what is the % yield of the reaction?
% yield = 1.38g / 2.77g x 100 = 49.8% % error = 2.77g g 2.77g X 100 = 50.2%  What is the relationship between percent yield and percent error?

71 How many grams of N2 are formed when 18
How many grams of N2 are formed when 18.1 grams of NH3 are reacted with 90.4 grams of CuO? 2NH3 + 3CuO  N2 + 3Cu 3H2O 18.1g 90.4g ?

72 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 28g N2 1mol N2
2NH3 + 3CuO  N2 + 3Cu + 3H2O 18.1g 90.4g ? 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 28g N2 1mol N2 = grams N2

73 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 28g N2 1mol N2
2NH3 + 3CuO  N2 + 3Cu + 3H2O 18.1g 90.4g ? 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 28g N2 1mol N2 = grams N2 90.4g CuO 1mol CuO 1mol N2 28g N2 79.55g CuO 3mol CuO = grams N2

74 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 = 14.91 grams N2 28g N2
2NH3 + 3CuO  N2 + 3Cu + 3H2O 18.1g 90.4g ? 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 = grams N2 28g N2 1mol N2 90.4g CuO 1mol CuO 1mol N2 28g N2 79.55g CuO 3mol CuO = grams N2

75 Theoretical amount of N2
2NH3 + 3CuO  N2 + 3Cu + 3H2O 18.1g 90.4g ? 18.1g NH3 1mol NH3 1mol N2 17g NH3 2mol NH3 = grams N2 28g N2 1mol N2 90.4g CuO 1mol CuO 1mol N2 28g N2 79.55g CuO 3mol CuO = grams N2 Theoretical amount of N2

76 CuO is the limiting reactant and we use it to see how much NH3 actually reacts:

77 CuO is the limiting reactant and we use it to see how much NH3 actually reacts:
90.4g CuO 1mol CuO 2mol NH3 17g NH3 79.55g CuO 3mol CuO 1mol NH3 = grams of NH3 reacts

78 = 12.88 grams of NH3 reacts 18.1g ­ 12.88g = 5.22g unreacted NH3
CuO is the limiting reactant and we use it to see how much NH3 actually reacts: 90.4g CuO 1mol CuO 2mol NH3 17g NH3 79.55g CuO 3mol CuO 1mol NH3 = grams of NH3 reacts 18.1g ­ 12.88g = 5.22g unreacted NH3

79 2NH3 + 3CuO  N2 + 3Cu + 3H2O 18.1g 90.4g 0g 0g 0g 12.88g 90.4g 10.61g ? ? 2NH3 + 3CuO  N2 + 3Cu + 3H2O 5.22g 0g 10.61g ? ?

80 9.48g / 10.61g x 100 = 89.35 % yield 10.61g – 9.48g = 1.13g = error
If the student only makes 9.48 grams of nitrogen gas, what is their percent yield? 9.48g / 10.61g x 100 = % yield What is the student’s absolute error and percent error in this experiment? 10.61g – 9.48g = 1.13g = error 1.13g / 10.61g x 100 = % error

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