2. 8 H : 8 O Imagine that paperclips are atoms and that we’re starting with an equal number of hydrogen atoms and oxygen atoms. For example: Let’s assemble our paperclip atoms into paperclip molecules. 2H 2 + O 2 -> 2H 2 O H2H2 O2O2

3. O2O2 H2H2 H2OH2O The product molecules are made out of the available reactant molecules.

4. O2O2 H2H2 H2OH2O Continue to convert reactants to products.

5. O2O2 H2H2 H2OH2O

6. O2O2 H2H2 H2OH2O

7. O2O2 H2H2 H2OH2O The reaction stops after forming 4 water molecules because there aren’t any more hydrogen molecules to react. The reaction was limited by the amount of hydrogen initially present. Therefore,

8. 2 H 2 + O 2  2 H 2 O O2O2 H2H2 + H2OH2O 4 mol H 2 4 mol O 2 Step 1: Assume all of the hydrogen reacts and determine the number of moles of water that would be formed. 4 mol H 2 O = 4 mol 8 mol H 2 O Step 2: Assume all of the hydrogen reacts and determine the number of moles of water that would be formed. = 8 mol Q: How many grams of water will be formed? Step 3: Multiply the smaller molar amount by the gfm of water to find the number of grams of water that will be formed. A: 4 mol x 18.015 g/mol 72 g H 2 O

9. 2 H 2 + O 2  2 H 2 O For the reaction below, how many moles of water will be formed if 1.55 moles of hydrogen gas is allowed to react with 0.30 moles of oxygen gas? What is the limiting reagent? 1.55 mol H 2 1.55 mol H 2 O 0.30 mol O 2 0.60 mol H 2 O Which reagent is in excess? How many grams of H 2 O (gfm = 18.015 g/mol) will be formed from this reaction? 0.60 mol H 2 O x 18.015 g/mol = 10.8 g of H 2 O will form

10. For the reaction below, how many grams of water will be formed if 1.17 g of hydrogen gas is allowed to react with 7.35 g of oxygen gas? What is the limiting reagent? 2 H 2 + O 2  2 H 2 O g: gfm: mol R1 : mol R2 : 1.17 g7.35 g 2.016 g/mol31.999 g/mol 0.58 mol ——0.23 mol 0.58 mol 0.46 mol x 18.015 g/mol 8.29 g How many grams of water will be formed? 8.29 g