1. Ch. 12 Stoichiometry 12.1 The Arithmetic of Equations

2. I. Interpreting Chem. Eqns. ___ N 2 (g) + ___ H 2 (g)  ___ NH 3 (g) 231 A. Balanced equations show how much product is made with the reactants B. Coefficients show relationship between reactants and products in particles, moles, and liters (if under same temperature and pressure conditions) Ex. 1 mole N 2 produces 2 moles of NH 3 if there is enough H 2 present Ex. 6 molecules of H 2 will produce 4 molecules of NH 3 if enough N 2 is present (3:2 ratio)

3. II. Sample Problems 1. ___CH 4 (g)+ ___O 2 (g)  ___CO 2 (g)+ ___H 2 O(g)2211 2. ___ particles of H 2 O produced from 4 particles of O 2 assuming enough CH 4 available? 4 3. ___ moles of CH 4 needed to make 6 moles of H 2 O assuming enough O 2 available? 3 4. ___ liters of H 2 O gas are made with 20 L of CO 2 gas40

4. 12.2 Chemical Calculations

5. I. Mole to Mole Ratio A. Can make conversion factor out of numbers in balanced equation B. ___ Al(s) + ___O 2 (g)  ___ Al 2 O 3 (s) 2324 C. ____ moles Al 2 O 3 made from 8 moles Al assuming enough O 2 ? 4 D. 8 moles Al = moles Al 2 O 3 Moles Al Moles Al 2 O 3 4 2 4

6. II. Other Conversions A. ___ H 2 (g) + ___ O 2 (g)  ___ H 2 O (g) 221 B. Convert 2.0 moles O 2  grams H 2 O with enough H 2 2.0 moles O 2 = grams H 2 O Moles O 2 Moles H 2 O 1 2 g H 2 O 1 18.02 72 C. 3.0 moles O 2  L H 2 O at STP with enough H 2 3.0 mol O 2 = L H 2 O Mole O 2 Moles H 2 O 1 2 Mole H 2 O L H 2 O 1 22.4 130

7. D. Convert 20.0 L H 2 at STP  grams O 2 20.0 L H 2 = grams O 2 L H 2 Mole H 2 22.4 1 Moles H 2 Moles O 2 2 1 Mole O 2 g O 2 1 32.00 14.3

8. 12.3 Limiting Reagent and Percent Yield

9. if not how many moles of salt could be made? ___ I. Limiting/Excess Reagents A. ___ Na + ___ Cl 2  ___ NaCl221 B. ___ moles NaCl made with 4 moles of Cl 2 assuming enough Na? 8 C. Will same amount of salt be possible if only 3 moles of Na? ___, No 3 D. When you start with 3 moles Na and 4 moles Cl 2 Na is the Limiting Reagent: limits amount of product E. Cl 2 is the Excess Reagent: it is left over

10. II. Identifying Limiting/Excess 1. Balance: ___ Cu + ___ S  ___ Cu 2 S 211 2. Start with: 4 moles Cu, 1 mole S Limiting:__________, Excess: __________ SulfurCopper 3. Start with: 3 moles Cu, 3 moles S Limiting:__________, Excess: __________ CopperSulfur 4. Start with: 80.0 grams Cu, 32.0 grams S Limiting:__________, Excess: __________ CopperSulfur

11. III. Limiting Calculation A. You can’t compare masses directly, change all values to moles first B. 2 Cu + 1 S  1 Cu 2 S 80.0 g Cu = moles Cu g Cu Moles Cu 63.5 1 1.26 C. 32.0 g S = moles S g S Moles S 32 1 1.0 D. Cu limiting because need at least 2 moles to match 1 mole S

12. IV. Percent Yield A. Percent Yield: mass of product you experimentally get compared to theoretical ideal mass B. % = (experimental mass/theoretical mass) x 100