1. Stoichiometry

2. The MOLE RATIO In a balanced chemical reaction, the coefficients tell you how many moles of each substance you need for the reaction and how many moles of each product you will produce (theoretically). C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 1mol C 3 H 8 + 5mol O 2  3mol CO 2 + 4mol H 2 O

3. What is the MOLE RATIO used for? The mole ratio is used to convert from one substance to another. (Dimensional Analysis style of course!) We use this because we don’t have any instruments that measure things in MOLES!

4. Example How many moles of CO 2 can theoretically be produced from 11.0 mol O 2 ? C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 11.0 mol O 2 3mol CO 2 = 6.60 mol CO 2 5mol O 2

5. PAUSE NOW TO TRY SOME EXAMPLES WITH YOUR FAVORITE CHEMISTRY TEACHER!

6. Of course, we don’t measure things in moles and most of the time our answers aren’t in moles either. We measure substances in… ◦ Mass (grams) ◦ Volume (mL) ◦ Molecules/Atoms Lucky for us, we know how to convert all of those into and out of moles!

7. 2HCl + CaCO 3  CO 2 + H 2 O + 2CaCl 2 How many moles of calcium chloride are produced from 100.0g of blackboard chalk, calcium carbonate? ◦ What are we given in the problem?  Balanced equation  100.0g of CaCO 3 ◦ What are we trying to find?  Moles of CaCl 2 ◦ What should we do 1 st ?

8. 2HCl + CaCO 3  CO 2 + H 2 O + 2CaCl 2 How many moles of calcium chloride are produced from 100.0g of blackboard chalk, calcium carbonate? 1 st – Convert your given into moles! ◦ 200.0g CaCO 3 1 mol CaCO 3 = 2.000mol 100.g CaCO 3 CaCO 3

9. 2HCl + CaCO 3  CO 2 + H 2 O + 2CaCl 2 How many moles of calcium chloride are produced from 200.0g of blackboard chalk, calcium carbonate? Then, take that answer and use it with your mole ratio! 2.000mol CaCO 3 2mol CaCl 2 = 4.000mol 1 mol CaCO 3 CaCl 2

10. 2FeCl 3 + 3H 2 S  Fe 2 S 3 + 6HCl 2FeCl 3 + 3H 2 S  Fe 2 S 3 + 6HCl What is the theoretical yield hydrochloric acid (in grams) produced when 8.53 × 10 24 molecules of hydrosulfuric acid react with excess iron (III) chloride? ◦ What are we given in the problem?  Balanced equation  8.53 × 10 24 molecules H 2 S ◦ What are we trying to find?  Mass of HCl produced

11. 2FeCl 3 + 3H 2 S  Fe 2 S 3 + 6HCl What is the theoretical yield hydrochloric acid (in grams) produced when 8.53 × 10 24 molecules of hydrosulfuric acid react with excess iron (III) chloride? 1 st – Convert those molecules to moles! 8.53 × 10 24 molecules H 2 S 1 mole H 2 S = 14.2 mol H 2 S 6.02 × 10 23 molecules H 2 S

12. 2FeCl 3 + 3H 2 S  Fe 2 S 3 + 6HCl What is the theoretical yield hydrochloric acid (in grams) produced when 8.53 × 10 24 molecules of hydrosulfuric acid react with excess iron (III) chloride? Now we can use the mole ratio! 14.2 mol H 2 S 6 mol HCl = 28.4 mol HCl 3 mol H 2 S Finally, we can turn those moles into mass! 28.4 mol HCl 36.5 g HCl = 1036.6g HCl = 1040g HCl 1 mol HCl