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Chemistry 20 Stoichiometry. This unit involves very little that is new. You will merely be applying your knowledge of previous units to a new situation.

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Presentation on theme: "Chemistry 20 Stoichiometry. This unit involves very little that is new. You will merely be applying your knowledge of previous units to a new situation."— Presentation transcript:

1 Chemistry 20 Stoichiometry

2 This unit involves very little that is new. You will merely be applying your knowledge of previous units to a new situation. The most important thing to remember is no matter what the problem is, follow the steps.

3 Stoichiometry Steps 1.Write and balance the chemical equation. 2.Determine what information is given and what you are required to calculate. 3. Covert the given information to moles. 4.Use the mole ratio to find the moles of unknown. 5. Covert the moles of unknown to the required unit.

4 Much of this unit will consist of writing examples in class. Please note that, while the steps are fundamental to solving problems in stoichiometry, they must be applied intelligently. For example, if you are given a balanced equation, then there is no need to balance it. Or if the given information is in moles then there is no need to convert to moles.Mole SongMole Song

5 Stoichiometry Stoichiometry is the study of chemical calculations. It begins with another look at a chemical equation. 2 H 2 O (l)  2 H 2 (g) + O 2 (g) We would read this as two molecules of water decomposes to two molecules of hydrogen gas and one molecule of oxygen gas.

6 The whole equation can be doubled: 4 H 2 O (l)  4 H 2 (g) + 2 O 2 (g) Or the whole equation can be written in terms of particles: 1.204 x 10 24 particles H 2 O (l)  1.204 x 10 24 particles H 2 (g) + 6.022 x 10 23 particles O 2 (g)

7 Or the whole equation can be written in terms of moles: 2 moles H 2 O (l)  2 moles H 2 (g) + 1 mole O 2 (g) Looking at the equation in terms of moles allows us to compare moles of one thing to moles of another.

8 Mole Ratio 2 H 2 O (l)  2 H 2 (g) + O 2 (g) How many moles of oxygen gas do you have compared to moles of water? You have 1 oxygen mole 2 water moles This is your mole ratio.

9 Mole Ratio Examples Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag How many moles of Cu to how many moles of AgNO 3 ? 1 Cu 2 AgNO 3

10 Mole Ratio Examples Continued Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag How many moles of Cu(NO 3 ) 2 to how many moles of AgNO 3 ? 1 Cu(NO 3 ) 2 2 AgNO 3

11 Using the Mole Ratio Go back to the water equation: 2 H 2 O (l)  2 H 2 (g) + O 2 (g) If you were asked how many moles of oxygen would be produced if you had 10 moles of water, you would use the mole ratio in a conversion. 1 oxygen = ? oxygen 2 water10 water Cross multiply (1)(10) = (2)(?) 10 =2(?) 5 = ? You would have 5 moles of oxygen gas.

12 Stoichiometry Example 1 2 H 2 + O 2  2 H 2 O How many moles of O 2 react to get 0.50 moles of water? Mole ratio 1 O 2 2 H 2 O Conversion and Cross Multiply 1 O 2 = ? O 2 2 H 2 O 0.50H 2 O (2)(?) = (1)(0.50) Answer 0.25 mole O 2

13 Stoichiometry Example 2 2 H 2 + O 2  2 H 2 O How many moles of O 2 react to with 16 moles of H 2 ? Mole ratio 1 O 2 2 H 2 Conversion and Cross Multiply 1 O 2 = ? O 2 2 H 2 16 H 2 (2)(?) = (1)(16) Answer 8.0 mole O 2

14 Stoichiometry Example 3 2 H 2 + O 2  2 H 2 O How many moles of O 2 react with 0.375 moles of H 2 ? Mole ratio 1 O 2 2 H 2 Conversion and Cross Multiply 1 O 2 = ? O 2 2 H 2 0.375 H 2 (2)(?) = (1)(0.375) Answer 0.188 mole O 2

15 Stoichiometry Example 4 N 2 + 3 H 2  2 NH 3 How many moles of each substance are present? N 2 1 mole H 2 3 moles NH 3 2 moles

16 Stoichiometry Example 5 N 2 + 3 H 2  2 NH 3 How many moles of H 2 are needed to get 0.50 moles of NH 3 ? Mole ratio 3 H 2 2 NH 3 Conversion and Cross Multiply 3 H 2 = ? H 2 2 NH 3 0.50 NH 3 (2)(?) = (3)(0.50) Answer 0.75 mole H 2

17 Up to this point we have been doing mole to mole calculations. You will need to know how to calculate grams to grams calculations as well.

18 Grams to Grams Questions

19 Grams to Grams Question 1 N 2 + 3 H 2  2 NH 3 How many grams of H 2 are required to form 42.5 grams of NH 3 ? First convert grams of known to moles 42.5 g NH 3 x 1 mole 17.03056 g 2.50 mol NH 3

20 Then continue like before… Mole ratio 3 H 2 2 NH 3 Conversion and Cross Multiply 3 H 2 = ? H 2 2 NH 3 2.50 NH 3 (2)(?) = (3)(2.50) Answer 3.75 mole H 2

21 Now convert moles back to grams 3.75 mol H 2 x 2.01588 g 1 mol H 2 7.56 grams of H 2

22 Grams to Grams Question 2 N 2 + 3 H 2  2 NH 3 How many grams of NH 3 are produced when 84.56 g of N 2 react? First convert grams of known to moles 84.56 g N 2 x 1 mole 28.01348 g 3.019 mol N 2

23 Then continue like before… Mole ratio 1 N 2 2 NH 3 Conversion and Cross Multiply 1 N 2 = 3.019 N 2 2 NH 3 ? NH 3 (2)(3.019) = (1)(?) Answer 6.038 mole NH 3

24 Now convert moles back to grams 6.038 mol NH 3 x 17.03056 g 1 mol 102.8 grams of NH 3

25 You now can calculate problems with Moles or Grams

26 There are other types of questions, but these will be done in Grade 12 Chemistry!


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