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STOICHIOMETRY. Recipe for Chocolate Cake: 2 c flour 1 c sugar 2 eggs 1 c oil ½ c cocoa X 3 6 c flour 3 c sugar 6 eggs 3 c oil 1½ c cocoa.

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Presentation on theme: "STOICHIOMETRY. Recipe for Chocolate Cake: 2 c flour 1 c sugar 2 eggs 1 c oil ½ c cocoa X 3 6 c flour 3 c sugar 6 eggs 3 c oil 1½ c cocoa."— Presentation transcript:

1 STOICHIOMETRY

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3 Recipe for Chocolate Cake: 2 c flour 1 c sugar 2 eggs 1 c oil ½ c cocoa X 3 6 c flour 3 c sugar 6 eggs 3 c oil 1½ c cocoa

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5 The calculations of quantities in chemical reactions is a subject of chemistry called stoichiometry.

6 Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or how much product will be formed in a reaction.

7 When you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction.

8 Numbers of atoms: 6 atoms H2 atoms N 6 atoms H 8 atoms = Conservation of matter 2 atoms N N N N N H H H H H H H H H HH H N 2 (g) + 3H 2 (g) 2NH 3 (g) ++

9 Numbers of moles: 3 moles H 2 2 moles NH 3 2 moles4 moles ≠ 1 mole N 2 N N N N H H H H H H H H H HH H N 2 (g) + 3H 2 (g) 2NH 3 (g) + # of moles are not conserved

10 Mass: 6 g 34 g = 28 g N N N N H H H H H H H H H HH H N 2 (g) + 3H 2 (g) 2NH 3 (g) + Conservation of Mass 14 + 143 X 2 14 + 3 = 17 2 X 17

11 Interpreting a Balanced Chemical Equation Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is: 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) + 2H 2 O(g)

12 Interpret this equation in terms of numbers of: b) representative particles. 2 molecules H 2 S3 molecules O 2 2 molecules SO 2 2 molecules H 2 O + +

13 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) + 2H 2 O(g) Interpret this equation in terms of numbers of: c) moles. 2 moles H 2 S 3 moles O 2 2 moles SO 2 2 moles H 2 O + +

14 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) + 2H 2 O(g) Interpret this equation in terms of: d) masses of reactants and products. 2 mol X 34.1 g/mol + 3 mol X 32.0 g/mol 2 mol X 64.1 g/mol 2 mol X 18.0 g/mol+ 68.2 g + 96.0 g 128.2 g 36.0 g + 164.2 g =

15 N 2 (g) + 3H 2 (g) 2NH 3 (g) MOLE RATIOS We know that: Based on this, you can write ratios that relate moles of reactants to moles of product: Mole Ratios:

16 N 2 (g) + 3H 2 (g) 2NH 3 (g) How many moles of NH 3 are produced when 0.60 moles of nitrogen reacts with hydrogen.

17 4 Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Mole ratios:

18 4 Al(s) + 3O 2 (g) 2Al 2 O 3 (s) How many moles of aluminum are needed to form 3.7 moles of Al 2 O 3 ?

19 4 Al(s) + 3O 2 (g) 2Al 2 O 3 (s) How many moles of oxygen are required to react completely with 14.8 moles of aluminum?

20 4 Al(s) + 3O 2 (g) 2Al 2 O 3 (s) How many moles of aluminum oxide are formed when 0.78 moles of oxygen reacts with aluminum?


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