Chemical Reaction - NSCC NetID Personal Web Space：化学反应和个人网络空间的用户名.pptVIP

* The setup would be: 13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2 18.0 g H2O 2 moles H2O 1 mole O2 molar mole-mole molar mass H2O factor mass O2 = 11.6 g O2 Calculating the Mass of a Reactant * Calculating the Mass of Product When 18.6 g ethane gas C2H6 burns in oxygen, how many grams of CO2 are produced? 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 18.6 g ? g The plan and factors would be g C2H6 mole C2H6 mole CO2 g CO2 molar mole-mole molar mass C2H6 factor mass CO2 * Calculating the Mass of Product The setup would be 18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2 30.1 g C2H6 2 moles C2H6 1 mole CO2 molar mole-mole molar mass C2H6 factor mass CO2 = 54.4 g CO2 * Active learning exercise * Lecture 4.0: Chemical Reactions and Quantities Percent Yield and Limiting Reactants Copyright ? 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings * Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product, which is calculated using the balanced equation. Actual yield The amount of product obtained when the reaction takes place. Percent yield The ratio of actual yield to theoretical yield. percent yield = actual yield (g) x 100 theoretical yield (g) * You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burn and you throw them out. The rest of the cookies are okay. What is the percent yield o