1. Stoichiometry Chapter 9 Limiting Reagents Stoich ppt _5 Limiting Reagents Practice

2. Stoichiometry After this presentation, you should understand:  Moles, mass, representative particles (atoms, molecules, formula units), molar mass, and Avogadro’s number.  Calculations using balanced chemical equations: for example, for a given mass of a reactant, calculate the volume of product.  Limiting reactants: calculate the mass of product formed when given the mass of each reactant.

3. Limiting Reagents Write the balanced chemical equation Write all quantitative values under equation (include units) Convert ALL reactants to moles (this is how much you HAVE) Do a mole-to-mole ratio using both reactants (this is how much you NEED) Compare HAVE to NEED to identify the limiting reactant!

4. Time to Practice! Grab your calculator and periodic table.

5. How many grams of silver can form when 12.0 g of copper are allowed to react with 15.0 g of silver nitrate? 1 Cu + 2 AgNO 3  2 Ag + 1 Cu(NO 3 ) 2 12.0 g? g 15.0 g HAVE:

6. How many grams of silver can form when 12.0 g of copper are allowed to react with 15.0 g of silver nitrate? 1 Cu + 2 AgNO 3  2 Ag + 1 Cu(NO 3 ) 2 0.188827694 mol ? g 0.088297621 mol HAVE NEED: Have > Need, so L.R. = AgNO 3 Use L.R. you HAVE to calculate product that can form

7. Time to Practice… extra challenging! Grab your calculator and periodic table.

8. Limiting Reactants The reaction between solid white phosphorus (P 4 ) and oxygen gas produces solid tetraphosphorous decoxide (P 4 O 10 ). a.Determine the mass of tetraphosphorous decoxide formed if 25.0 g of phosphorous (P 4 ) and 50.0 g of oxygen are combined. b.How much of the excess reactant remains after the reaction stops?

9. How do we solve this? 1.What do we know? mass of phosphorous = 25.0 g P 4 mass of oxygen = 50.0 g O 2 2.Write a balanced equation for the reaction 1 P 4 (s) + 5O 2 (g)  1 P 4 O 10 (s) 25.0 g50.0 g a) ? g b) ? excess

10. Determine the number of moles of reactants available 1 P 4 (s) + 5O 2 (g)  1 P 4 O 10 (s) 25.0 g P 4 x 1 mol P 4 = 0.202 mol P 4 1123.88 g P 4 HAVE 50.0 g O 2 x 1 mol O 2 = 1.56 mol O 2 1 32.00 g O 2 HAVE 25.0 g50.0 g

11. 1 P 4 (s) + 5 O 2 (g) ---  1 P 4 O 10 (s) Calculate the moles of O 2 needed to react completely with the available moles of P 4. 0.202 moles P 4 x 5 moles O 2 = 1.01 mol O 2 11 mol P 4 NEED

12. Compare the moles O 2 needed to the moles O 2 available: Need:Have: 1.01 mol O 2 <1.56 mol O 2 Because more moles of O 2 are available than are needed, the O 2 is in excess and P 4 is the limiting reactant.

13. Use the limiting reactant’s moles to determine how much product will be produced. 0.202 mol P 4 x 1 mol P 4 O 10 = 0.202 mol P 4 O 10 1 1 mol P 4 To find the mass: 0.202 mol P 4 O 10 x 283.9 g P 4 O 10 = 57.3 g P 4 O 10 1 1 mol P 4 O 10

14. b. How much of the excess reactant remains after the reaction goes to completion? ** Difference between HAVE & NEED ** HAVE = 50.0 g O 2 & NEED = 1.01 mol Convert so units match! 50.0 g O 2 (have) – 32.32 g O 2 (need) 17.7 g O 2 in excess