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Understanding Stoichiometry in the Catalytic Oxidation of Ammonia for Nitric Acid Production

This document explores the stoichiometry involved in the catalytic oxidation of ammonia (NH3) during the industrial production of nitric acid. The key reaction, which is NH3(g) + O2(g) → NO(g) + H2O(g), is analyzed while using 824 g of NH3 with excess oxygen. The reaction is first balanced to facilitate the calculation of moles produced. We focus on determining how many moles of nitrogen monoxide (NO) and water (H2O) are formed as products. This process underlines the critical role of stoichiometry in chemical manufacturing.

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Understanding Stoichiometry in the Catalytic Oxidation of Ammonia for Nitric Acid Production

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  1. Stoichiometry • Part 3: mass to moles

  2. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.

  3. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g)

  4. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)

  5. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen.

  6. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed?

  7. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed?

  8. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? NH3(g) + O2(g) → NO(g) + H2O(g)

  9. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? NH3(g) + O2(g) → NO(g) + H2O(g) First, we balance the reaction.

  10. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? NH3(g) + O2(g) → NO(g) + 3 H2O(g) First, we balance the reaction.

  11. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? 2 NH3(g) + O2(g) → NO(g) + 3 H2O(g) First, we balance the reaction.

  12. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? 2 NH3(g) + O2(g) → 2 NO(g) + 3 H2O(g) First, we balance the reaction.

  13. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? 2 NH3(g) + 2½ O2(g) → 2 NO(g) + 3 H2O(g) First, we balance the reaction.

  14. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? 2( 2 NH3(g) + 2½ O2(g) → 2 NO(g) + 3 H2O(g) ) First, we balance the reaction.

  15. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction.

  16. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol.

  17. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol.

  18. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol.

  19. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol.

  20. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol.

  21. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol.

  22. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value.

  23. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value.

  24. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value.

  25. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values.

  26. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values.

  27. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values.

  28. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values.

  29. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy.

  30. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy. 1. Calculate the number of mols of NH3.

  31. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy. 1. Calculate the number of mols of NH3.

  32. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy. 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO.

  33. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy. 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO.

  34. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy. 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  35. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ First, we balance the reaction. Second, we write down the molar masses, Ṃ, in g/mol. Third, we write the known value. Fourth, we write down the unknown values. Fifth, we plan our solution strategy. 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  36. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  37. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  38. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  39. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  40. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  41. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  42. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  43. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. 3. Calculate the number of mols of H2O.

  44. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia 3. Calculate the number of mols of H2O.

  45. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia ➠ nNO = (nammonia)(coeffNO)/coeffammonia 3. Calculate the number of mols of H2O.

  46. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia ➠ nNO = (nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 3. Calculate the number of mols of H2O.

  47. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia ➠ nNO = (nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol 3. Calculate the number of mols of H2O.

  48. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 ? ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia ➠ nNO = (nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol 3. Calculate the number of mols of H2O.

  49. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 48.4 ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia ➠ nNO = (nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol 3. Calculate the number of mols of H2O.

  50. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)The reaction is run using 824 g of NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed? Ṃ (g/mol) 17.04 32.00 20.01 18.01 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) m (g) 824 n (mol) 48.4 48.4 ? ⓵ ⓶ ⓷ 1. Calculate the number of mols of NH3. nammonia = mammonia/Ṃammonia = (824 g)/(17.04 g/mol) = 48.4 mol 2. Calculate the number of mols of NO. nNO/nammonia = coeffNO/coeffammonia ➠ nNO = (nammonia)(coeffNO)/coeffammonia ➠ nNO = (48.4 mol)(4)/4 = 48.4 mol 3. Calculate the number of mols of H2O.

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