Chapter 12: Stoichiometry Limiting Reagent and Percent Yield

Learning Target You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed. You will calculate the percent yield of a reaction.

Limiting and Excess Reagents You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Limiting Reagent Problem Example Copper reacts with sulfur to form copper (I) sulfide according to the following balanced equation: 2 Cu (s) + S (s)  Cu2S (s) What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S? You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

How to solve Pick a given amount to start with and perform a stoichiometric calculation to figure out theoretical yield of other reactant. Compare theoretical yield to given amount of reactant If theoretical yield > given yield, then LR If theoretical yield < given yield, then ER (make sure units are same) Use LR as given value and solve as we’ve been You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Shortcut (ONLY for LR) Convert the amount of each reactant to moles (if not already) Divide each mole amount by its respective coefficient Smallest number is the limiting reactant You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #1 Copper reacts with sulfur to form copper (I) sulfide according to the following balanced equation: 2 Cu (s) + S (s)  Cu2S (s) What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S? 100. g Cu2S You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #1 (work) 2 Cu (s) + S (s)  Cu2S (s) What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S? 100. g Cu2S You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #2 The heat from an acetylene torch is produced by burning acetylene (C2H2) in oxygen. 2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g) How many grams of water can be produced by the reaction of 2.40 mol C2H2 with 7.40 mol O2? 43.2 g H2O You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g) Example #2 (work) 2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g) How many grams of water can be produced by the reaction of 2.40 mol C2H2 with 7.40 mol O2? 43.2 g H2O You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Try on your own Aluminum reacts with chlorine gas to form aluminum chloride according to the balanced equation 2 Al (s) + 3 Cl2 (g)  2 AlCl3 (s) In a certain experiment, 10.0 g of aluminum is reacted with 35.0 g of chlorine gas. What mass of aluminum chloride will be produced, assuming a complete reaction? = 43.9 g AlCl3 You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Try on your own 1st-Find the LR 2 Al (s) + 3 Cl2 (g)  2 AlCl3 (s) 10.0 g 35.0 g ? g 10.0 g Al x 1 mol Al x 3 mol Cl2 x 70.90 g Cl2 1 26.98gAl 2molAl 1molCl2 = 39.4 g Cl2 NEEDED…> 35.0 g Cl2 GIVEN, LR You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Try on your own (cont.) 2nd-Find the mass of AlCl3 2 Al (s) + 3 Cl2 (g)  2 AlCl3 (s) 10.0 g 35.0 g ? g 35.0 gCl2 x 1molCl2 x 2molAlCl3 x133.33 gAlCl3 1 70.90 g Cl2 3 mol Cl2 1molAlCl3 = 43.9 g AlCl3 You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #2 Lithium nitride, an ionic compound containing the Li+ and N3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium in the unbalanced reaction Li (s) + N2 (g)  Li3N (s) = 93.7 g Li3N You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g 56.0 g ? g You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed. Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g 56.0 g ? g 1. Pick a given amount 56.0 g Li of a reactant: 1 2. Convert that 56.0 g Li x 1 mol Li amount to moles: 1 6.941 g Li

Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g 56.0 g ? g You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed. Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g 56.0 g ? g 3. Mole ratio to convert to moles of other reactant 56.0 g Li x 1 mol Li x 1 mol N2 1 6.941 g Li 6 mol Li

*this mass is the amount needed in the reaction Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g 56.0 g ? g 4. Convert that mole amount to mass 56.0 g Li x 1 mol Li x 1 mol N2 x 28.02g N2 1 6.941 g Li 6 mol Li 1 mol N2 = 37.7 g N2  mass needed of N2 in reaction *this mass is the amount needed in the reaction You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g 56.0 g ? g = 37.7 g N2  mass calculated of N2 needed in reaction 5. Compare the mass needed to the mass given in the problem 37.7 g = mass calculated (needed); 56.0 g = mass given 37.7 g < 56.0 g ; calculated < given so excess reactant You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Example #2 6 Li (s) + N2 (g)  2 Li3N (s) 56.0 g (LR) ? g 56.0 g Li x 1 mol Li x 2 mol Li3N x 34.833 g Li3N 1 6.941 g Li 6 mol Li 1 mol Li3N = 93.7 g Li3N You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) Try on your own Iron (III) oxide reacts with carbon monoxide to form iron metal and carbon dioxide gas: Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) In a certain experiment, 5.0 g of iron (III) oxide is reacted with 5.0 g of carbon monoxide gas. What mass of iron will be produced, assuming a complete reaction? = 3.5 g Fe You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) Try on your own 1st-find LR Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) 5.0 g 5.0 g ? g 5.0gFe2O31molFe2O3 x 3molsCOx28.01gCO 1 159.7gFe2O3 1molFe2O3 1molCO = 2.6 g CO NEEDED…< 5.0 g GIVEN, ER You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) Try on your own 2nd –find mass of Fe Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) 5.0 g 5.0 g ? g 5.0gFe2O31molFe2O3 x 2molsFe x 55.85g Fe 1 159.7gFe2O3 1molFe2O3 1molFe = 3.5 g Fe You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed.

Percent Yield What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated? CaCO3 (s)  CaO (s) + CO2 (g) You will calculate the percent yield of a reaction.

Example What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated? CaCO3 (s)  CaO (s) + CO2 (g) 94.2% You will calculate the percent yield of a reaction.

CaCO3 (s)  CaO (s) + CO2 (g) Example (work) What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated? CaCO3 (s)  CaO (s) + CO2 (g) 94.2% You will calculate the percent yield of a reaction.

SiO2 (s) + 3 C (s)  SiC (s) + 2 CO (g) Try on your own If 5.00 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO2 (s) + 3 C (s)  SiC (s) + 2 CO (g) What is the percent yield of this reaction? 83.5% You will calculate the percent yield of a reaction.

SiO2 (s) + 3 C (s)  SiC (s) + 2 CO (g) Try on your own (work) If 5.00 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO2 (s) + 3 C (s)  SiC (s) + 2 CO (g) What is the percent yield of this reaction? 83.5% You will calculate the percent yield of a reaction.

You will calculate the percent yield of a reaction. Quiz Consider the following chemical reaction: 3 H2 + N2  2 NH3. If you are given 6 molecules of H2 and 4 molecules of N2, what is the limiting reactant? 2 KCl + 3 O2  2 KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2, what is the limiting reactant? Explain what a limiting reactant is to a student who has been absent from this class for a few days. If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is produced. What is the percent yield of this reaction? (use the equation in #1) You will determine the limiting reactant in a problem, the theoretical yield, and the amount of excess reactant remaining when the reaction is completed. You will calculate the percent yield of a reaction.

You will learn to understand the concept of a limiting reactant, recognize it in a reaction, and use it to solve stoichiometric calculations. Homework Page 284 34ac and 38