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Stoichiometry Chapter 8. Stoichiometry Chemical equations Limiting reagent Problem types Percent yield Mass-mass Mole - mole other.

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Presentation on theme: "Stoichiometry Chapter 8. Stoichiometry Chemical equations Limiting reagent Problem types Percent yield Mass-mass Mole - mole other."— Presentation transcript:

1 Stoichiometry Chapter 8

2 Stoichiometry Chemical equations Limiting reagent Problem types Percent yield Mass-mass Mole - mole other

3 Ways to Interpret Chemical Equations 1.Representative Particles example: O 2 is two oxygen atoms or 1 molecule of oxygen 2.Moles Example: 2O 2 is 2 mol of Oxygen 3.Mass Example: 2O 2 is 2(16 x 2) = 64 grams of oxygen. 4.Volume

4 Mole-Mole Calculations N 2 (g) + 3H 2 (g) 2NH 3 (g) 1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia. How many moles of ammonia are produced when.6 mol of nitrogen reacts with hydrogen? Calculate using a mole ratio

5 Mole-Mole Calculations, cont..6 mol N 2 x 2 mol NH 3 /1 mol N 2 = (.6 mol N 2 )(2 mol NH 3 )/ 1 mol N 2 =.6 x 2 mol NH 3 = 1.2 mol NH 3

6 Practice Problem Calculate the number of moles of nitrogen and hydrogen that are required to make 7.24 mol of ammonia. – Remember the balanced equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) – Set up mole ratios – Cancel units accordingly – Multiply

7 Mass – Mass Calculations N 2 (g) + 3H 2 (g) 2NH 3 (g) -see example 6 on pg. 216 ***easiest to set up labels and cancel***


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