Limiting Reactants and Percent Yield Honors Chemistry

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.

Excess The reactant that you have more than you need to perform a chemical reaction is called the excess reactant It is not completely used up in a chemical reaction.

Here’s an example You are making ham and cheese sandwiches and you have: 5 pieces of ham 5 pieces of cheese 8 pieces of bread Which of these ingredients do you have more than enough of (excess reactant)? Which of these do you not have enough of (limiting reactant?

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. Subtract that value from the initial mass of the excess reactant.

1. What mass 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 reactant 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 2.02 g H2 24.31g Mg 1mol Mg 1mol H2 = 4.15 gH2 75.0 g HCl 1mol HCl 1 mole H2 2.02 g H2 36.46 g HCl 2mol HCl 1mol H2 = 2.08 g H2 HCl is the limiting reactant!!

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!

2. What mass of calcium sulfate can be produced from the reaction of 1000 g calcium phosphate with 980 g H2SO4? 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 and 1317g of calcium phosphate are produced

Percent Yield Percent Yield describes 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

Write balanced reaction Cu + S  Cu2S Example 1: 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

Example 2: A chemist starts with 1 Example 2: A chemist starts with 1.75 g of salicylic acid (C7H6O3) and excess methanol (CH3OH) and reports the production of 1.42 g oil of wintergreen (C8H8O3) in the following reaction. What is the percent yield for this reaction? C7H6O3 + CH3OH  C8H8O3 + H2O 1. 1.75 g ?g 2. 1.75 g x 1 mol = 138 g 0.0127 mol C7H6O3 3. 0. 0127 mol C7H6O3 x 1 molC8H8O3 = 1 mol C7H6O3 0.0127 mol C8H8O3 4. 0.0127 mol C8H8O3 x 152 g = 1 mol 1.93 g C8H8O3 5. (1.42 g ÷ 1.93 g) x 100 = 73.6 % Yield

Practice Problems A chemist carried out a reaction that should produce 21.8 g of a product, according to a mass-mass calculation. However, the chemist was able to recover only 13.9 g of the product. What percentage yield did the chemist get? A calculation indicates that 82.2 g of a product should be obtained from a certain reaction. If a chemist actually gets 30.7 g, what is the percentage yield? Chromium(III) hydroxide will dissolve in sodium hydroxide according to the following equation: NaOH + Cr(OH)3  NaCr(OH)4 If you begin with 66.0 g of Cr(OH)3 and obtain 38.4 g of product, what is your % yield?

Solutions #1. (13.9g / 21.8g) x 100 = 63.8% #2. (30.7g / 82.2g) x 100 = 37.3% #3. NaOH + Cr(OH)3  NaCr(OH)4 66.0g ?g 66.0g Cr(OH)3 x 1mol/103g = 0.641mol Cr(OH)3 = 0.641mol NaCr(OH)4 x 143g / 1mol NaCr(OH)4 = 91.66g NaCr(OH)4 (38.4g / 91.7g) x 100 = 41.9%

Practice Problem Identify the limiting reactant and the theoretical yield of H3PO3 if 225 g of PCl3 is mixed with 125 g of H2O PCl3 + 3H2O  H3PO3 + 3HCl Convert each mass to moles: 225 g PCl3 x 1 mol/137 g = 1.64 mol PCl3 125 g H2O x 1 mol/ 18 g = 6.94 mol H2O 1.64 mol PCl3 requires 4.92 mol H2O PCl3 is the limiting reactant 1.64 mol PCl3 = 1.64 mol H3PO3 1.64 mol H3PO3 x 82 g/1 mol = 134 g H3PO3 225g 125g ?g