2. 1 Mr. ShieldsRegents Chemistry Unit 11 L03

3. 2 Once we’ve balanced a chemical equation What other information does if provide? For example: What information does the following give us? N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) OK. Let’s see … Hydrazine Hydrogen peroxide

4. 3 N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) This equation provides the following information: 1. What’s reacting and what’s produced 2. The states of matter involved in the reaction 3. How many molecules of each reactant are needed for the reaction 4. How many molecules of product are produced If we know how many molecules are involved then We can also state this equation in terms of moles. So, Why is that?

5. 4 2 molecules 1 molecule 2 molecules Which is: 2 x N A 1 x N A 2 x N A. Let’s look at the reaction between H 2 and O 2 It’s just a question of “scale”. Whether it’s 2 molecules to 1 molecule or 2 moles to 1 mole

6. 5 In or prior example 1 mole of hydrazine plus 2 moles of hydrogen peroxide yield 1 mole of Nitrogen and 4 moles of water. N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) If we react these chemicals together in the Laboratory must we insure we have exactly 1 mole of of hydrazine and 2 moles of hydrogen Peroxide present ? No. Let’s see why…

7. 6 Whether we talk about 1 molecule or 6.023 x 10 23 molecules (1 mole) or even 3.01 x 10 23 It’s just a question of “scale” We only need to insure the ratio’s remain the same Remember… we said we can state chemical equations in terms of molecules or in terms of number of moles.

8. 7 So what would happen if I double the # of Moles of reactants? N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) 2N 2 H 4 (g) + 4H 2 O 2 (l)  xN 2 (g) + xH 2 0 (l) Right … I double the # of moles of product Why? 2N 2 H 4 (g) + 4H 2 O 2 (l)  2N 2 (g) + 8H 2 0 (l) We must keep all the Mole ratio’s the same!

9. 8 In this simple problem let’s see how we would calculate the new mole ratio’s. What is the ratio of N 2 H 4 to N 2 and N 2 H 4 to H 2 0 in the original balanced eqn.? N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) So, 2N 2 H 4 (g) + 4H 2 O 2 (l)  2N 2 (g) + 8H 2 0 (l) 1:1 And 1:4 Lets look at Nitrogen first: If hydrazine is inc. to 2 mol N 2 H 4 N 2 Then 1 : 1x = 2 2 : x For water: If hydrazine is inc. to 2 then 1 : 4 x = 8 2 : x

10. 9 If I cut the number of moles of reactants in half How many moles of product will be produced? N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) ½ N 2 H 4 (g) + 1H 2 O 2 (l)  ½ N 2 (g) + 2 H 2 0 (l) Right … the number of moles of the products Are also reduced by half. (what are the mole ratios involved?) Let’s see how we would use this in solving Some problems.

11. 10 What would you predict the number of moles of Reactant and products to be when 6 moles of Hydrazine reacts completely with H 2 O 2 ? 6N 2 H 4 (g) + xH 2 O 2 (l)  xN 2 (g) + xH 2 0 (l) 1N 2 H 4 (g) + 2H 2 O 2 (l)  1N 2 (g) + 4H 2 0 (l) The ratio of N 2 H 4 to H 2 O 2 is 1:2 or 6:12 The ratio of N 2 H 4 to N 2 is 1:1 or 6: 6 The ratio of N 2 H 4 to H 2 O is 1:4 or 6:24 OK… 1 st you need to recall the balanced equation Then we need to look at the change in mole ratios

12. 11 So our balanced equation changes from N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) 6 N 2 H 4 (g) + 12 H 2 O 2 (l)  6 N 2 (g) + 24 H 2 0 (l) To OK. That’s easy, right? Let’s try another one

13. 12 How many moles of N 2 H 4 would be required to Completely react 0.25 moles of hydrogen Peroxide? b) How man moles of product would be formed? Remember … first you need to know what the Balanced equation is. N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) x N 2 H 4 (g) + ¼ H 2 O 2 (l)  x N 2 (g) + x H 2 0 (l)

14. 13 N 2 H 4 (g) + 2H 2 O 2 (l)  N 2 (g) + 4H 2 0 (l) x N 2 H 4 (g) + ¼ H 2 O 2 (l)  x N 2 (g) + x H 2 0 (l) The ratio of H 2 O 2 to N 2 H 4 is 2:1 or 0.25 : 0.125 The ratio of H 2 O 2 to N 2 is 2:1 or 0.25: 0.125 The ratio of H 2 O 2 to H 2 O is 1:2 or 0.25: 0.5 Let’s look at the change in mole ratios of H 2 O 2 to each of the other reactant & Product molecules So what is the final equation?.125 N 2 H 4 (g) + 0.25 H 2 O 2 (l) .125 N 2 (g) + 0.5 H 2 0 (l)

15. 14 A Word Problem: A chemist wants to produce 2.5 mol of water by reacting Hydrogen with Oxygen. How many moles Of each will he need? 2H 2 + O 2  2H 2 O (balanced eqn) Ratio of H 2 0 to H 2 = 1:1 Ratio of H 2 0 to O 2 = 2:1 2.5 mol of H 2 and 1.25 mol of O 2