Limiting Reagents, Theoretical, Actual and Percent Yields

University:
California State University, Northridge
Course:
CHEM 101 | General Chemistry I
Academic year:

2023
Views:

483

Pages:

3
Author:

Eddie M.

LIMITING REAGENTS, THEORETICAL , ACTUAL AND PERCENT YIELDS A limiting reagent is a chemical reactant that limits the amount of product that is formed. The limiting reagent gives the smallest yield of product calculated from the reagents (reactants) available. This smallest yield of product is called the theoretical yield. To find the limiting reagent and theoretical yield, carry out the following procedure: 1. Find the moles of each reactant present. 2. Calculate the moles of a product formed from each mole of reactant. 3. Identify the reactant giving the smaller number of moles of product. This reactant is the Limiting Reagent: 4. Calculate the grams of product produced by the Limiting Reagent. This is the theoretical yield. THEORETICAL YIELD The theoretical yield is the amount of the product in g formed from the limiting reagent. From the moles of limiting reagent available, calculate the grams of product that is theoretically possible (same as Step 4 above). ACTUAL YIELD The actual yield is the amount of the product in g actually formed in the laboratory. PERCENT YIELD The percent yield is the percent of the product formed based upon the theoretical yield. actual yield in g ----------------------- x 100 % theoretical yield in g = Percent Yield 1 LIMITING REAGENTS, THEORETICAL , ACTUAL AND PERCENT YIELDS EXAM PLE OF A LIM ITING REAGENT PROBLEM How many grams of NH 3 can be be produced (theoretically) from the reaction of 5.0 g of N 2 and 5.0 H 2? What is the limiting reagent? If 8.52 g are actually formed, what is the percent yield of NH 3? 3H 2 Balance: N + 1 mole H 2: 2 x 1.01 g H = 2.01 g Find: 2 NH3 ” 2 1 mole N 2: 2 x 14.01 g = 28.02 g 1 mole NH 3: 3 x 1.01 g 1 x 14.01 g = 3.03 g H = 14.01 g N -----------17.04 g NH3 To answer this type of question, carry out EACH step of the following procedure: 1. Find the g of product produced from each of the given amounts of reactants. For H 2: 5.0 g H 2 x For N 2 : 5.0 g N 2 1 mole H 2 2.02 g H2 x1 mole N 2 140.01 g N2 x 2 mole NH 3 x 3 mol H 2 x 2 mole NH 3 1 mol N 2 x 17.04 g NH 3 1 mol NH 3 17.04 g NH 3 1 mol NH 3 = 28.12 g NH 3 = 12.16 g NH 3 (carry 1 or 2 additional significant figures here) 2. Identify the smaller amount of calculated product: 12.16 g is the smaller amount calculated. This is the maximum amount of product that can be formed. The larger amount cannot be formed because there is not enough of the limiting reagent present. Identity the reactant that gives the smallest amount of product. The limiting reagent determines the amount of product formed. 3. The limiting reagent is N 2. 12 g is the theoretical yield 4. Identify the THEORETICAL YIELD: The theoretical yield is the amount of the product (in g) formed from the limiting reagent. 8.25 g is the actual yield. 5. Identify the ACTUAL YIELD: The actual yield is the amount of the product (in g or mol) actually formed in the laboratory. 6. Calculate the PERCENT YIELD: The percent yield is based upon the theoretical yield. actual yield (g) -----------------------x 100 % theoretical yield (g) = 8.25 g = ---------- x 100 % = 12.16 g Percent Yield 68% (For theoretical yield, carry more sig. figs. in this calculation) Final sig. figs. limited by the given amounts of reactants. 2