1. WARM UP The reaction of iron ore with carbon follows the equation: 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2. How many moles of Fe can be produced from 37 moles Fe 2 O 3 and 35 moles of C?

2. Reaction Amounts Chemistry II

3. Yield Gram method – Conversions – Ex: 2 CO (g) + O 2(g)  2 CO 2(g) 50 grams CO will yield _______ g CO 2

4. Theoretical Yield Maximum mass of products from given reactants – Need balanced equation (moles) and gram/mole conversions Oxygen can be produced by the decomposition of mercury(II) oxide, through the following balanced reaction: 2 HgO(s)  2 Hg(l) + O 2 (g). How many grams of O 2 will be produced by the reaction of 24.2 g of HgO?

5. Actual Yield Measured amount actually obtained – In grams Always less than theoretical yield – Poor recovery, incomplete reaction, contamination

6. Percent Yield Efficiency of reaction Actual yield Theoretical yield x 100 In the reaction of 4.0 moles of N 2 with 6.0 moles of H 2, a chemist obtained 1.6 moles of NH 3. What is the percent yield of NH 3 ? (The balanced equation is 3 H 2 + N 2  2 NH 3 ).

7. Limiting and Excess Reagents Using grams – Mole/gram conversions – Determine yields based on each reactant Reactant making fewest grams of product is limiting Excess left over based on amount given and amount needed to react with the limiting reagent

8. WRAP UP Calculate the amount of Fe you can produce from 409 g of Fe 2 O 3. 2 Fe 2 O 3  4 Fe + 3 O 2

9. WARM UP How much H 2 do you need to completely react with 26.8 g of O 2 in order to form water? 2 H 2 + O 2  2 H 2 O

10. WRAP UP The reaction of iron ore with carbon follows the equation: 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2. – How many grams of Fe can be produced from a mixture of 200 g of Fe 2 O 3 and 300 g of C? – In the above problem, which reactant is limiting and which is in excess? – How much of the excess reagent will remain after the reaction is complete?