The document provides examples and practice problems for calculating percent yield in chemical reactions. It defines theoretical yield, actual yield, and percent yield. Sample problems show how to use these concepts to calculate percent yield when given the actual and theoretical yields. Further practice problems ask the reader to calculate percent yields or actual yields given various amounts in chemical equations.
4. Read the more challenging sample problem below. Sample Problem: 16 g of H 2 reacts with excess O 2 to produce 138 g of H 2 O. What is the percent yield for this chemical reaction? Step 1: Write the balanced chemical equation (if it isn’t already given to you). 2H 2 + O 2 2H 2 O Step 2: Determine the actual and theoretical yield. The actual yield is given in the problem. Actual yield = 138 g of H 2 O
5. The theoretical yield is calculated (use Mass- Mass Stoichiometry from your arrow sheet). Continued… Step 3: Calculate the percent yield. The theoretical yield is 143 g of H 2 O. 16 g H 2 % yield = x 100% x 100% 1 mol H 2 2.02 g H 2 x 2 mol H 2 O 2 mol H 2 x 18.02 g H 2 O 1 mol H 2 O x 143 g = actual theoretical 138 g H 2 O 1 4 3 g H 2 O = 96.7% =
![[object Object],[object Object],[object Object],[object Object],[object Object]](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)