1. More Stoichiometry Problems involving Volume or Particles or Energy Chapter 9 Section 1, pages 308 – 311

2. Stoichiometry Strategy All problems involve three basic steps: 1.Get to moles of ‘given’ 2.Use the stoichiometric ratio 3.Get out of moles of ‘find’ Units given Units find mol given mol find coeff. find coeff. given Which means we just need to figure out how to convert between moles and units of measure.

3. Gas Volume Problems (Using Molar Volume: 22.4 L/mol) Example: What volume of NH 3 forms when 672 mL H 2 completely reacts? N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Vol given Vol find mol given mol find coeff. find coeff. given √

4. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Step 1 – get to moles of ‘given’ (note that volume must be converted to liters) 672 mL H 2 1 mol H 2 22.4 L 0.0300 mol H 2 Vol given Vol find mol given mol find coeff. find coeff. given √ √ 1 L 1000 mL Gas Volume Problems (Using Molar Volume: 22.4 L/mol)

5. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Step 2 – Use the stoichiometric ratio: 0.0200 mol NH 3 2 mol NH 3 3 mol H 2 0.0300 mol H 2 Vol given Vol find mol given mol find coeff. find coeff. given √ √√

6. Gas Volume Problems (Using Molar Volume: 22.4 L/mol) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Step 3 – get out of moles of ‘find’ (This requires two conversions if you want the answer in milliliters): Vol given Vol find mol given mol find coeff. find coeff. given √ √√ √ 0.0200 mol NH 3 1000 ml 1 L 448 mL NH 3 22.4 L NH 3 1 mol NH 3

7. Gas Volume Problems (Using Molar Volume: 22.4 L/mol) Putting all steps together we have: Vol L of given Vol L of find mol given mol find coeff. find coeff. given 1 3 2 672 mL H 2 1 L 1000 mL 1 mol H 2 22.4 L H 2 2 mol NH 3 3 mol H 2 22.4 L NH 3 1 mol NH 3 1000 ml 1 L 448 mL NH 3 x molar volume  molar volume

8. Particle Stoichiometry Problems Example: How many molecules of C 5 H 8 form from 1.89 x 10 24 molecules C 5 H 12 ? C 5 H 12 (l)  C 5 H 8 (l) + 2 H 2 (g) molecules given molecules find mol given mol find coeff. find coeff. given √

9. Particle Stoichiometry Problems C 5 H 12 (l)  C 5 H 8 (l) + 2 H 2 (g) Step 1 – get to moles of given: 1.89 x 10 24 molecules C 5 H 12 1 mol 6.022 x 10 23 3.14 mol C 5 H 12 molecules given molecules find mol given mol find coeff. find coeff. given √ √  Avogadro’s number

10. Particle Stoichiometry Problems C 5 H 12 (l)  C 5 H 8 (l) + 2 H 2 (g) Step 2 – Use the stoichiometric ratio: 3.14 mol C 5 H 8 1 mol C 5 H 8 1 mol C 5 H 12 3.14 mol C 5 H 12 molecules given molecules find mol given mol find coeff. find coeff. given √ √√

11. Particle Stoichiometry Problems C 5 H 12 (l)  C 5 H 8 (l) + 2 H 2 (g) Step 3 – get out of moles of ‘find’: 1.89 x 10 24 molecules C 5 H 8 6.022 x 10 23 1 mol C 5 H 8 3.14 mol C 5 H 8 molecules given molecules find mol given mol find coeff. find coeff. given √ √ √ √ x Avogadro’s number

12. Particle Stoichiometry Problems Putting all steps together we have: 1.89 x 10 24 molecules C 5 H 8 1 mol C 5 H 8 1 mol C 5 H 12 1.89 x 10 24 molecules C 5 H 12 1 mol 6.022 x 10 23 1 mol C 5 H 8 molecules given molecules find mol given mol find coeff. find coeff. given  Avogadro’s number 1 32 x Avogadro’s number

13. Coefficient Find Coefficient Given moles given moles find particles given grams given volume L of given particles find grams find volume L of find  6.022 x 10 23 X 6.022 x 10 23 X molar mass  molar mass The Mole Road  molar volume X molar volume