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Stoichiometry Mole-Mole Problems

Stoichiometry Mole-Mole Problems. Mr. Shields Regents Chemistry Unit 11 L03. 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).

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Stoichiometry Mole-Mole Problems

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  1. Stoichiometry Mole-Mole Problems Mr. Shields Regents Chemistry Unit 11 L03

  2. Once we’ve balanced a chemical equation What other information does if provide? For example: What information does the following give us? N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) Hydrazine Hydrogen peroxide OK. Let’s see …

  3. N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) • This equation provides the following information: • What’s reacting and what’s produced • The states of matter involved in the reaction • 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?

  4. 2 molecules 1 molecule 2 molecules Which is: 2 x NA 1 x NA 2 x NA. Let’s look at the reaction between H2 and O2 It’s just a question of “scale”. Whether it’s 2 molecules to 1 molecule or 2 moles to 1 mole

  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. N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (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…

  6. Remember… we said we can state chemical equations in terms of molecules or in terms of number of moles. Whether we talk about 1 molecule or 6.023 x 1023 molecules (1 mole) or even 3.01 x 1023 It’s just a question of “scale” We only need to insure the ratio’s remain the same

  7. So what would happen if I double the # of Moles of reactants? N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) 2N2H4(g) + 4H2O2 (l)  xN2 (g) + xH20 (l) Right … I double the # of moles of product Why? 2N2H4(g) + 4H2O2 (l)  2N2 (g) + 8H20 (l) We must keep all the Mole ratio’s the same!

  8. In this simple problem let’s see how we would calculate the new mole ratio’s. What is the ratio of N2H4 to N2 and N2H4 to H20 in the original balanced eqn.? N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) 1:1 And 1:4 Lets look at Nitrogen first: If hydrazine is inc. to 2 mol N2H4 N2 Then 1 : 1 x = 2 2 : x For water: If hydrazine is inc. to 2 then 1 : 4 x = 8 2 : x So, 2N2H4(g) + 4H2O2 (l)  2N2 (g) + 8H20 (l)

  9. If I cut the number of moles of reactants in half How many moles of product will be produced? N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) ½ N2H4(g) + 1H2O2 (l)  ½ N2 (g) + 2 H20 (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.

  10. What would you predict the number of moles of Reactant and products to be when 6 moles of Hydrazine reacts completely with H2O2? 6N2H4(g) + xH2O2 (l)  xN2 (g) + xH20 (l) OK… 1st you need to recall the balanced equation 1N2H4(g) + 2H2O2 (l)  1N2 (g) + 4H20 (l) Then we need to look at the change in mole ratios The ratio of N2H4 to H2O2 is 1:2 or 6:12 The ratio of N2H4 to N2 is 1:1 or 6: 6 The ratio of N2H4 to H2O is 1:4 or 6:24

  11. So our balanced equation changes from N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) To 6 N2H4(g) + 12 H2O2 (l)  6 N2 (g) + 24 H20 (l) OK. That’s easy, right? Let’s try another one

  12. How many moles of N2H4 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. N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) x N2H4(g) + ¼ H2O2 (l)  x N2 (g) + x H20 (l)

  13. N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l) x N2H4(g) + ¼ H2O2 (l)  x N2 (g) + x H20 (l) Let’s look at the change in mole ratios of H2O2 to each of the other reactant & Product molecules The ratio of H2O2 to N2H4 is 2:1 or 0.25 : 0.125 The ratio of H2O2 to N2 is 2:1 or 0.25: 0.125 The ratio of H2O2 to H2O is 1:2 or 0.25: 0.5 So what is the final equation? .125 N2H4(g) + 0.25 H2O2 (l)  .125 N2 (g) + 0.5 H20(l)

  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? 2H2 + O2 2H2O (balanced eqn) Ratio of H20 to H2 = 1:1 Ratio of H20 to O2 = 2:1 2.5 mol of H2 and 1.25 mol of O2

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