1. STOICHIOMETRY CALCULATIONS COACH COX

2. MOLE TO MOLE CONVERSIONS Converting from moles of one substance in a chemical reaction to moles of another substance in a chemical reaction

3. MOLE TO MOLE: EXAMPLE 1 If you dropped 0.040 moles of potassium into a beaker of water, how many moles of hydrogen would you produce? Step 1:Write a balanced chemical equation. ____K (s) + ____ H 2 O (l)  ____ KOH (aq) + ____ H 2(g) Step 2:Start with your known, and use Dimensional Analysis with the correct mole ratio from the balanced equation to find the unknown - Draw your pathway!

4. MOLE TO MOLE: EXAMPLE 1 ANSWER Balanced Equation: 2K (s) + 2 H 2 O (l)  2 KOH (aq) + 1 H 2(g) 1 mol H 2 Pathway:0.040 mol K ------------------------------------> moles of H 2 2 mol K 0.040 mol K x 1 mol H 2 =.02 mol H 2 produced 2 mol K

5. MOLE TO MOLE: EXAMPLE 2 Using the same number of moles of potassium, how many moles of H 2 O will be used? Balanced Equation: 2K (s) + 2 H 2 O (l)  2 KOH (aq) + 1 H 2(g) Work out the question the exact same way you did Example 1!

6. MOLE TO MOLE: EXAMPLE 2 ANSWER Balanced Equation: 2K (s) + 2 H 2 O (l)  2 KOH (aq) + 1 H 2(g) 2 mol H 2 O Pathway:0.040 mol K ------------------------------------> moles of H 2 O 2 mol K 0.040 mol K x 2 mol H 2 O =.04 mol H 2 O will be used 2 mol K

7. MOLE TO MOLE: EXAMPLE 3 One disadvantage of burning propane, C 3 H 8, is that CO 2 is produced as a product. The released CO 2 increases the growing concentration of carbon dioxide in the atmosphere. How many moles of CO 2 are produced when 10.45 moles of C 3 H 8 is burned in a propane gas grill? ___ C 3 H 8(g) + ___ O 2(g)  ___ CO 2(g) + ___ H 2 O (l) Work it out the same way, balance the equation first!

8. MOLE TO MOLE: EXAMPLE 3 ANSWER Balanced Equation: 1 C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) 3 mol CO 2 Pathway:10.45 mol C 3 H 8 ------------------------------------> moles of CO 2 1 mol C 3 H 8 10.45 mol C 3 H 8 x 3 mol CO 2 = 31.35 mol CO 2 will be produced 1 mol C 3 H 8

9. MOLE TO MOLE: EXAMPLE 4 The reaction between methane and, CH 4, and sulfur produces carbon disulfide, CS 2, a liquid used in the production of cellophane. ___CH 4 + ___S 8  ___CS 2 + ___H 2 S a.Balance the equation. b.Calculate the moles of CS 2 produced when 1.78 moles of S 8 is consumed. c.How many moles of H 2 S is produced during this same reaction?

10. MOLE TO MOLE: EXAMPLE 4 ANSWER a. Balanced Equation:2 CH 4 + 1 S 8  2 CS 2 + 4 H 2 S b. 1.78 mol S 8 x 2 mol CS 2 = 3.56 mol CS 2 will be produced 1 mol S 8 c. 1.78 mol S 8 x 4 mol H 2 S = 7.12 mol H 2 S will be produced 1 mol S 8

11. MOLE TO MASS CONVERSIONS Converting from moles of one substance in a chemical reaction to mass of another substance in a chemical reaction

12. MOLE TO MASS EXAMPLE 1 Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts with sodium Step 1:Write a balanced equation ___ Cl 2 + ___ Na  ___ NaCl Step 2: Start with your known, and use Dimensional Analysis with the correct mole ratio from the balanced equation to find the unknown - Draw your pathway!

13. MOLE TO MASS: EXAMPLE 1 ANSWER Balanced Equation:1 Cl 2 + 2 Na  2 NaCl Pathway:1.25 mol Cl 2 ------------------------------------------> grams NaCl produced 1.25 mol Cl 2 x 2 mol NaCl x 58.443 g NaCl = 146.11 grams NaCl produced 1 mol Cl 2 1 mol NaCl

14. MOLE TO MASS EXAMPLE 2 Determine the mass of the sodium needed to complete this same reaction. Balanced Equation:1 Cl 2 + 2 Na  2 NaCl Work out the question the exact same way you did Example 1!

15. MOLE TO MASS: EXAMPLE 2 ANSWER Balanced Equation:1 Cl 2 + 2 Na  2 NaCl Pathway:1.25 mol Cl 2 ------------------------------------------> grams Na needed 1.25 mol Cl 2 x 2 mol Na x 22.990 g Na = 57.475 grams Na needed 1 mol Cl 2 1 mol Na

16. MASS TO MASS CONVERSIONS Converting from the mass of one substance in a chemical reaction to the mass of another substance in a chemical reaction

17. MASS TO MASS CONVERSIONS EVERYTHING MUST GO THROUGH MOLES!!! (MOLE TO MOLE conversions)

18. MASS TO MASS EXAMPLE 1 Determine the mass of H 2 O produced from the decomposition of 25.0 g of ammonium nitrate (NH 4 NO 3 ) Step 1:Write a balanced equation ___ NH 4 NO 3  ___ H 2 O + ___ N 2 O Step 2: Start with your known, and use Dimensional Analysis to convert from: a. mass of known to moles of known using the periodic table b. moles of known to moles or unknown using the correct mole ratio moles to mass of the unknown using the periodic table - Draw your pathway!

19. MASS TO MASS: EXAMPLE 1 ANSWER Balanced Equation: 1 NH 4 NO 3  2 H 2 O + 1 N 2 O Pathway:25 g NH 4 NO 3 ---------------------------------------> grams H 2 O produced 25 g NH 4 NO 3 x 1 mol NH 4 NO 3 x 2 mol H 2 O x 18.015 g H 2 O = 97.829 g NH 4 NO 3 1 mol NH 4 NO 3 1 mol H 2 O = 9.207 grams of H 2 O produced

20. MASS TO MASS EXAMPLE 2 Calculate the number of grams of NH 3 produced by the reaction of 5.4 g of hydrogen with an excess of nitrogen ___ N 2(g) + ___ H 2(g)  ___ NH 3(g)

21. MASS TO MASS: EXAMPLE 2 ANSWER Balanced Equation: 1 N 2(g) + 3 H 2(g)  2 NH 3(g) Pathway:5.4 g H 2 ---------------------------------------> grams NH 3 produced 5.4 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17.031 g NH 3 = 2.016 g H 2 1 mol H 2 1 mol NH 3 = 91.2375 grams of NH 3 produced

22. PARTICLES TO PARTICLES CONVERSIONS Converting from the particles of one substance in a chemical reaction to the particles of another substance in a chemical reaction

23. PARTICLES TO PARTICLES CONVERSIONS EVERYTHING MUST GO THROUGH MOLES!!! (MOLE TO MOLE conversions)

24. PARTICLES TO PARTICLES CONVERSIONS: EXAMPLE 1 Sodium hydrogen carbonate decomposes with heat to produce carbon dioxide gas, water, and solid sodium carbonate. 2.2 x 10 23 formula units of sodium hydrogen carbonate react to form ________ molecules of carbon dioxide. ___ NaHCO 3(s)  ___ Na 2 CO 3(s) + ___ CO 2(g) + ___ H 2 O (g)

25. PARTICLES TO PARTICLES CONVERSIONS: EXAMPLE 1ANSWER Balanced Equation:2 NaHCO 3(s)  1 Na 2 CO 3(s) + 1 CO 2(g) + 1 H 2 O (g) Pathway: 2.2 x 10 23 fun NaHCO 3 ---------------------------------------> molecules CO 2 2.2 x 10 23 fun NaHCO 3 x 1 mol NaHCO 3 x 1 mol CO 2 x 6.02x10 23 molecules CO 2 = 6.02x10 23 fun NaHCO 3 2 mol NaHCO 3 1 mol CO 2 = 1.1 x 10 23 molecules of CO 2 produced