Determining Limiting Reactant - Sandwiches and Stoichiometry

1. Starter A sandwich consists of two slices of bread, 3 slices of meat, and one slice of cheese. For each of the following amounts, determine the number of sandwiches that can be made and what is left over: • 6 bread, 10 meat, 4 cheese slices • 10 bread, 6 meat, 8 cheese slices • 25 bread, 40 meat, 12 cheese slices

2. Starter • 6 bread, 10 meat, 4 cheese slices • 3 sandwiches • 0 bread, 1 meat, 1 cheese • 10 bread, 6 meat, 8 cheese slices • 2 sandwiches • 6 bread, 0 meat, 6 cheese • 25 bread, 40 meat, 12 cheese slices • 12 sandwiches • 1 bread, 4 meat, 0 cheese

3. Ch. 9 Stoichiometry 9.3 Limiting Reactant

4. Why is there a limiting reactant? • a reaction rarely has exactly the right amount of each reactant • usually have some left over • limiting reactant • reactant that limits the amount of product created • always completely used up • excess reactant • reactant not completely used up

5. When do you have to find a LR? • whenever two amounts of reactants are given in a problem • when only one amount of reactant is given in a problem, then the other is assumed to be in excess

6. Finding Limiting Reactant using reactants • Figure out how much you need of B if you use up all of A • Convert grams A to grams B using stoichiometry • You may start with either reactant • Determine whether you will have enough • If you don’t have enough of B, then B is LR • If you don’t have enough of A, then A is LR

7. Finding Limiting Reactant using products • Convert each of the reactant amounts into an amount of product (doesn’t matter which product) • Compare product amounts and find lowest amount. • Whichever reactant led to lowest product amount is LR

8. Example 1 • The reaction begins with 2.51 g of HF and 4.56 g of SiO2. What is the limiting reactant and the excess reactant? How much excess reactant will be left over? • Write the balanced chemical equation SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)

9. Example 1 • Find the number of moles available of each reactant:

10. Example 1 • If we use up all of the HF, how much SiO2 will we need to go with it? • Do we have enough SiO2? • 0.0759 mol available > 0.0313 mol needed • YES- there will be some left over • Limiting Reactant : HF

11. How much of the product can be formed? • Start conversion with amount of limiting reactant. • Convert to amount of product using stoichiometry

12. Example 1 • How many grams of water could be formed? • Convert grams of HF to moles. • Convert moles of HF to moles of water. • Convert moles to grams using molar mass.

13. Example 2 • A reaction was done with 36.8 g C6H6 and 41.0 g of O2. • Write the balanced chemical equation 2C6H6 + 15O2 12CO2 + 6H2O • What is the limiting reactant and how much of each product can be produced?

14. Example 2 Reactant Method 36.8 g C6H6 1 mol C6H6 15 mol O2 32 g O2 78.12 g C6H6 2 mol C6H6 1 mol O2 = 113 g O2 Because 113 g O2 is greater than what we have available, O2 is LR Product Method 36.8 g C6H6 1 mol C6H6 6 mol H2O 18.02 g H2O 78.12 g C6H6 2 mol C6H6 1 mol H2O = 25.5 g H2O 41.0 g O2 1 mol O2 6 mol H2O 18.02 g H2O 32 g O2 15 mol O2 1 mol H2O = 9.24 g H2O Because 9.24 is less than 25.5, O2 is the LR

15. Example 2 • Once you find the LR, you can then calculate the amount of each product formed… Amount of H2O 41.0 g O2 1 mol O2 6 mol H2O 18.02 g H2O 32 g O2 15 mol O2 1 mol H2O = 9.24 g H2O Amount of CO2 41.0 g O2 1 mol O2 12 mol CO2 44.01 g CO2 32 g O2 15 mol O2 1 mol CO2 = 45.11 g CO2

16. Example 3 • If the reaction below begins with 51.03 grams of Fe and 37.5 grams of oxygen, what is the limiting reactant? 4Fe(s) + 3O2(g)  2Fe2O3

17. Example 3 • How many grams of oxygen will be left over after the reaction? • How many grams of iron (III) oxide can be formed?