Limiting Reactants and Percent Yield

1. Limiting Reactants and Percent Yield AP Chemistry

2. What is a Limiting Reactant? • It is the reactant in a reaction that determines how much product can be made. • It is whatever reactant you have the least amount of. • If you are making a bicycle and you have all the parts to make 100 bikes, but only 4 wheels available, how many bikes can you make? • What is the limiting “part”? • For chemistry, it is whatever has the least amount of moles.

3. Use the steps below to solve the following problem to determine the limiting reactant. 1. Write a balanced equation. 2. Do a separate mass to mass problem starting with each reactant. The smaller answer is correct. To find out how much of the excess reactant is left over, 1. Start with the initial mass of the limiting reactant and 2. Do a mass to mass problem to determine how much of the excess reactant was needed. 3. Subtract that value from the initial mass of the excess reactant.

4. 1. What volume of hydrogen gas at STP is produced from the reaction of 50.0g of Mg and 75.0 grams of HCl? How much of the excess reagent is left over (in grams)? Mg(s) + HCl(aq)  MgCl2(s) + H2(g) 2 Do a standard mass to mass problem starting with each reactant 50.0 gMg 1mol Mg 1mol H2 22.4 L 24.31g Mg 1mol Mg 1mol H2 = 46.1 L H2 75.0 g HCl 1mol HCl 1 mole H2 22.4 L 36.46 g HCl 2mol HCl 1mol H2 = 23.0 L H2 HCl is the limiting reactant!!

5. How much of the excess reactant is left over? • Start with the initial mass of the limiting reactant and do a mass to mass with the other reactant. • 75.0g HCl 1mol HCl 1mol Mg 24.31g Mg 36.46g HCl 2mol HCl 1mol Mg = 25.0 grams Mg needed for the reaction Thus, 50.0 grams – 25.0 grams = 25.0 grams Mg leftover!

6. 2. What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)? Ca3(PO4) 2 (s) + H2SO4(aq)  CaSO4(s) + H3PO4(aq) 3 3 2 1000g Ca3(PO4)2 1mol Ca3(PO4)2 3mol CaSO4 136.15g CaSO4 310.18g 1 mol Ca3(PO4)2 1mol CaSO4 = 1317 g 980 g H2SO4 1mol H2SO4 3mol CaSO4 136.15g CaSO4 98.09g H2SO4 3mol H2SO4 1mol CaSO4 = 1360 g • Calcium phosphate is the limiting reactant. • Starting with calcium phosphate, a mass to mass problem gives 632 grams phosphoric acid will form!

7. Percent Yield • Percent Yielddescribes how much product was actually made in the lab versus the amount that theoretically could be made. • Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made. • Theoretical Yield = The maximum amount of product that could be formed from given amounts of reactants. (you get this from doing a mass to mass Stoichiometry calculation!) • Actual Yield = The amount of product actually formed or recovered when the reaction is carried out in the laboratory. • % Yield = Actual Yield X 100 Theoretical Yield

8. Example - When copper is heated with an excess of sulfur, copper(I)sulfide is formed. In a given experiment, 1.50 g copper was heated with excess sulfur to yield 1.76 g copper(I) sulfide. What is the theoretical yield? What is the percent yield? Write balanced reaction Cu + S  Cu2S 2 Determine theoretical yield – doing a mass to mass problem 1.50g Cu 1 mol Cu 1mol Cu2S 159.17g Cu2S 63.55g Cu 2 mol Cu 1mol Cu2S = 1.88 g Cu2S Percent Yield = 1.76 g x 100 = 93.6 % 1.88g