Chemical Reactions and the Mole….

1. Chemical Reactions and the Mole…. 1 Fe (s) + 1 S (s) 1 FeS (s) This chemical reaction means one atom of iron reacts with one atom of sulfur to form one molecule of iron (II) sulfide Can we do this chemical reaction? No - atoms are too small to see - We must perform the reaction many times larger to see it! We always perform the reaction 6.02 x 1023 times larger, or 1 mole times larger

2. 1 Fe (s) + 1 S (s) 1 FeS (s) If we dothis reaction 6.02 x 1023 times bigger, it would be written: 6.02 x 1023 Fe (s) + 6.02 x 1023 S (s) 6.02 x 1023 FeS (s) Since 6.02 x 1023 is 1 mole, we could also say: 1 mole Fe (s) + 1 mole S (s) 1 mole FeS (s) So we could view the 1 coefficient as either 1 atom, or 1 mole, depending on whether we were looking at the reaction on the atomic scale, or the large, human, molar scale!

3. 1 Fe (s) + 1 S (s) 1 FeS (s) Here’s the problem - If you have 10 grams of Fe, how many grams of S do you need? Most would say 10 grams! It is a 1:1 ratio! But the units in the chemical reaction aren’t grams! The 1 coefficient does not stand for grams - it stands for moles! If you are going to Mexico, your dollars have to be converted into pesos before you go! When you look at the ratio in a chemical reaction, you must be in moles!

4. Let’s look at a sample problem… Number one in your homework packet under Easy Stoichiometry Problems… Carbon dioxide can be commercially prepared by heating chalk, or calcium carbonate. How many moles of carbon dioxide can be produced when 3.05 moles of calcium carbonate are heated? How do we approach this problem? There are four steps we follow in any problem we do with chemical reactions and amounts…

5. Write a balanced chemical reaction Make sure you are in the unit of moles (mole map) Set up a ratio from the chemical reaction, putting what you want to solve for on top, and what you want to cancel on bottom Convert out of the unit of moles, if necessary (mole map)

6. Stoichiometry Concept Map Volume of A (liquid) Volume of B (liquid) USE YOUR “CHEMICAL” RECIPE HERE Use Molarity Use Molarity moles Liter moles Liter X moles you want moles you have Use Molar Mass Use Molar Mass Moles A Moles B Grams A Grams B SET UP A RATIO FROM YOUR BALANCED EQUATION STP STP Use 22.4 L 1 mole Use 22.4 L 1 mole Volume of A (gas) Volume of B (gas)

8. Step 1: Write a balanced chemical equation: 1 CaCO3 (s) 1 CaO (s) + 1 CO2 (g) Step 2: Convert into the unit of moles: Done! We already have 3.05 moles CaCO3 Step 3: Set up a ratio: The reaction says that I can create 1 mole of CO2 for every 1 mole of CaCO3 I start with It is just like a recipe!

9. 1 CaCO3 (s) 1 CaO (s) + 1 CO2 (g) 3.05 moles CaCO3 x 1 mole CO2 = 1 mole CaCO3 3.05 moles CaCO3 No step 4 is required, because the problem asks for the answer in moles!

10. Let’s look at problem #10…. Group 1 metals are explosive when they come into contact with water. If 20.0 grams of potassium were to explode with excess water, how many moles of hydrogen would be produced? Let’s write a balanced chemical equation - step 1! 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H2 (g) Writing reactions will be CRITICAL on the test!!!

11. 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H2 (g) Step 2: Convert into moles: 20.0 g K x 1 mole = .51 moles K 39.1 grams I MUST be in the unit of the mole to be able to look at the ratios or amounts from the chemical reaction! Step 3: Set up a ratio from the reaction: .51 moles K x 1 mole H2 = .255 moles H2 2 mole K No step 4 is required because the problem asks for moles of hydrogen

12. Let’s look at problem #29…. Cigarette lighters use butane, or C4H10, as their fuel. If 120 grams of butane burn with only 55 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant? Let’s write a balanced chemical equation - step 1! 2 C4H10(g) + 13 O2(g) 10 HOH (g) + 8 CO2 (g) There is a problem here…. One of our starting chemicals is going to run out!

13. 2 C4H10(g) + 13 O2(g) 10 HOH (g) + 8 CO2 (g) To figure out which starting material runs out, called our limiting reactant, we must first convert both into moles: 120.0 g C4H10 x 1 mole = 2.07 moles C4H10 58 grams 55 L O2 x 1 mole = 2.46 moles O2 22.4 liters • Which one runs out? • Is it always the one we have less of? • What if I was making sandwiches, and used two pieces of bread, and one piece of bologna, per sandwich? • With 8 pieces of bologna, and 10 pieces of bread, what runs out….?

14. The bread! • I need twice as much bread as bologna to make a sandwich…. • In this problem, what runs out?

15. 2 C4H10(g) + 13 O2(g) 10 HOH (g) + 8 CO2 (g) 120.0 g C4H10 x 1 mole = 2.07 moles C4H10 58 g C4H10 55 L O2 x 1 mole O2 = 2.46 moles O2 22.4 L O2 To see which one runs out, we divide each number by the number needed in the reaction… 2.07/2 for the C4H10, and 2.46/13 for the O2 Which of these numbers is less? That is the one that runs out… The O2 is less, so it runs out! We finish the problem with the amount of O2, and ignore the C4H10 All problems are done this way - finish the problem with the chemical that runs out, and ignore the one you have extra of!

16. 2 C4H10(g) + 13 O2(g) 10 HOH (g) + 8 CO2 (g) Step 3: Set up a ratio from the reaction: 2.46 moles O2 x 8 moles CO2 = 1.51 moles CO2 13 moles O2 Step 4: Convert out of the mole: 1.51 moles CO2 x 22.4 L = 33.85 L CO2 1 mole CO2 Now try some problems on your own!