Stoichiometry and Percent Yields
Review:
know how to
Stoichiometric Problems
Consider the reaction of C6H6 + Br2 -> C6H5Br + HBr
a. What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2?
42.1 g of C6H6 x (1 mol of C6H6/(78.1134g of C6H6)) x (1 mol of C6H5Br/(1 mol of C6H6)) x (157.0095g of C6H5Br/(1 mol of C6H5Br)) = 84.6g of C6H5Br
73.0g of Br2 x (1 mole of Br2/(159.808g of Br2)) x (1 mol of C6H5Br/(1 mol of Br2)) x (157.0095g of C6H5Br/(1 mol of C6H5Br)) = 71.7g of C6H5Br
theoretical yield of C6H5Br is Br2
b. If the actual yield is 63.6g, what is the percent yield?
63.6/71.7 x 100 = 88.7%
Zn + S -> ZnS
a. If 25.0g of zinc and 30.0g of sulfur are reacted, what chemical is the limiting reagent?
25.0g of Zn x (1 mol of Zn/(65.39g of Zn)) x (1 mol of ZnS/(1 mol of Zn)) x (97.455g of ZnS/(1 mol of ZnS)) = 37.3g of ZnS
30.0g of S x (1 mol of S/(32.065g of S)) x (1 mol of ZnS/(1 mol of S)) x (97.455g of ZnS/(1 mol of ZnS)) = 91.2g of ZnS
Zn is the limiting reagent
b. How many grams of the excess reactant is left over after the reaction?
91.2-37.3 = 53.9g of ZnS x (1 mol of ZnS/(97.455g of ZnS)) x (1 mol of S/(1 mol of ZnS)) x (32.065/(1 mol of S)) = 17.7g of S
know how to
- go from name to compound
- balance equations
- predict products
- recognize combustion, double displacement, and single displacement
- solubility rules
Stoichiometric Problems
Consider the reaction of C6H6 + Br2 -> C6H5Br + HBr
a. What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2?
42.1 g of C6H6 x (1 mol of C6H6/(78.1134g of C6H6)) x (1 mol of C6H5Br/(1 mol of C6H6)) x (157.0095g of C6H5Br/(1 mol of C6H5Br)) = 84.6g of C6H5Br
73.0g of Br2 x (1 mole of Br2/(159.808g of Br2)) x (1 mol of C6H5Br/(1 mol of Br2)) x (157.0095g of C6H5Br/(1 mol of C6H5Br)) = 71.7g of C6H5Br
theoretical yield of C6H5Br is Br2
b. If the actual yield is 63.6g, what is the percent yield?
63.6/71.7 x 100 = 88.7%
Zn + S -> ZnS
a. If 25.0g of zinc and 30.0g of sulfur are reacted, what chemical is the limiting reagent?
25.0g of Zn x (1 mol of Zn/(65.39g of Zn)) x (1 mol of ZnS/(1 mol of Zn)) x (97.455g of ZnS/(1 mol of ZnS)) = 37.3g of ZnS
30.0g of S x (1 mol of S/(32.065g of S)) x (1 mol of ZnS/(1 mol of S)) x (97.455g of ZnS/(1 mol of ZnS)) = 91.2g of ZnS
Zn is the limiting reagent
b. How many grams of the excess reactant is left over after the reaction?
91.2-37.3 = 53.9g of ZnS x (1 mol of ZnS/(97.455g of ZnS)) x (1 mol of S/(1 mol of ZnS)) x (32.065/(1 mol of S)) = 17.7g of S