1. Quantities in Chemical Reactions (Part 1) or Stoichiometry

2. You’re a car maker now! Here’s the balanced “equation” for making a car: 1 body + 4 wheels + 2 wipers  1 chem- mobile Try this:

3. 1 body + 4 wheels + 2 wipers  1 chem- mobile How many chem-mobiles can you make from 35 bodies? 35. How many chem-mobiles can you make using 80 wheels? 20. How many chem-mobiles can you make using 98 wiper blades? 49.

4. What assumption(s) are you making? 1 body + 4 wheels + 2 wipers  1 chem-mobile For 98 wiper blades, we assume that we have at least _______ bodies. 49; and _______ wheels. 196. The “coefficients” in this “equation” are the equivalent of moles in a balanced chemical equation.

5. Let’s take this to chemistry... Li 3 N(s) + 3H 2 O(l)  NH 3 (aq) + 3 LiOH(aq) How many mol of lithium nitride are required to prepare 6 mol LiOH? 2 mol Li 3 N. Assume sufficent H 2 O. How many mol of water are required to react with 0.5 mol Li 3 N? 1.5 mol H 2 O.

6. Look at ratios in this equation... Li 3 N(s) + 3H 2 O(l)  NH 3 (aq) + 3 LiOH(aq) in terms of moles: 1 mol3 mol1 mol 3 mol in terms of mass: 34.8 g54.1 g17.0 g71.9 g in terms of molecules: 6.02E2318.06E236.02E2318.06E23 formula unitsmoleculesmoleculesformula units

7. Li 3 N(s) + 3H 2 O(l)  NH 3 (aq) + 3 LiOH(aq) What mass of NH 3 is expected from the reaction of 100 g of Li 3 N with sufficient water?

8. Li 3 N(s) + 3H 2 O(l)  NH 3 (aq) + 3 LiOH(aq) 1 mol Li 3 N : 1 mol NH 3 100 g ↓(/34.82 g·mol -1 ) 2.87 mol  (x 1/1)  2.87 mol ↓(x 17.0 g·mol -1 ) 48.8 g NH 3 expected

9. What mass of LiOH is expected from the reaction of 100 g of Li 3 N with sufficient water? 1 mol Li 3 N :3 mol LiOH 100 g ↓(/34.82 g·mol -1 ) 2.87 mol  (x 3/1)  8.62 mol ↓(x23.95 g·mol -1 ) 206 g LiOH expected

10. If 78.3 g of water is allowed to react with excess lithium nitride, how many molecules of ammonia would be produced? 3 mol H 2 O:1 mol NH 3 78.3 g ↓(/18.02 g·mol -1 ) 4.35 mol  (x 1/3)  1.45 mol ↓ (x 6.02E23molec·mol -1 ) 8.73E23 molecules NH 3 expected

11. Try this one: What volume of H 2 (g) at SATP can be obtained from the reaction of 8.54 g of Mg with sufficient HCl(aq)? Mg(s) + 2 HCl(aq)  MgCl 2 (aq) + H 2 (g) 8.54 g ↓(/24.3 g·mol -1 ) 0.351 mol  (x 1/1)  0.351 mol H 2 ↓[ x24.8L·mol -1 (SATP)] 8.70 L H 2 @ SATP expected

12. Get busy on this here work before I hog-tie y’all: p 298 PP 1 – 10 (do a few) p 300 PP 11 – 18 (do odd #s) p 304 PP 21 – 30 (do odd #s) p 305 RQ 4 – 10 (do as many as needed) Quiz next class