1. Limiting Reactants Unit 7 1

2. Limiting Reactants 2

3. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) If we start off with 100 moles of N 2 and 100 moles of H 2, how many moles of NH 3 can be produced? Reaction Number Moles N 2 leftMoles H 2 left Moles NH 3 produced 1 2 25 33.33 50

4. Limiting Reactants 4 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) If we start off with 100 moles of N 2 and 100 moles of H 2 :

5. Limiting Reactants 5 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) If we start off with 100 moles of N 2 and 100 moles of H 2 :

6. Limiting Reactants 6 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) If we start off with 100 moles of N 2 and 100 moles of H 2 :

7. Limiting Reactants 7 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) If we start off with 100 moles of N 2 and 100 moles of H 2, how many moles of NH 3 can be produced? How much reactants are left?

8. Limiting Reactants 8 Cu(s) + H 2 SO 4 (aq)  CuSO 4 (aq) + H 2 O(l) + SO 2 (g) How much SO 2 (in grams) can be produced from 18.00 g of Cu and 18.00 g of H 2 SO 4 ? How much copper is left over?

9. Limiting Reactants 9 Aluminum oxide can be produced from the following chemical reaction How much Al 2 O 3 can be prepared (in grams) from 50.0 g of Al and 100.0 g of O 2 ? 4 Al + 3 O 2  2 Al 2 O 3 10.0 g of Al and 10.0 g of O 2 11.0 g of Al and 9.40 g of O 2

10. Theoretical and Percent Yield 10 Theoretical Yield Maximum amount of a product that can be produced from a given amount of reactants Actual Yield The amount of a product actually obtained from a chemical reaction Percent Yield = Actual Yield Theoretical Yield 50.0 g of Al and 100.0 g of O 2 4 Al + 3 O 2  2 Al 2 O 3 If the Percent Yield is 82%, how much Al 2 O 3 is actually produced?