Question

Model 4: Using Moles of Limiting Reagent in Stoichiometry The above pictures are not realistic because they involve so very few molecules. In a "real-world" sample, you are likely to have on the order of 10 molecules. Consider the following: Suppose you have a reaction vessel initially containing 13.5 moles of sulfur dioxide and 11.2 moles of oxygen mol SO, required to react with all O2 = 11.2 mol O2 x 2 mol SO 22.4 mol SO, 1 mol O2 Since there aren't 22.4 moles of sulfur dioxide available, we conclude that sulfur dioxide is the limiting reagent and oxygen is the reagent in excess. The numbers of moles of sulfur trioxide and left over oxygen when the reaction is complete are: 2 mol so, mol So; produced - 13.5 mol SO, X = 13.5 mol SO, 2 mol SO, 1 mol O2 mol O, consumed - 13.5 mol SO, X -6.75 mol O2 2 mol SO, mol O2 remaining - 11.2 mol O2 initially - 6.75 mol O2 consumed - 4.45 mol O2 Key Questions 3a. (Circle the correct answer. It is the number of moles of the excess reagent or limiting reagent that is used to determine how many moles of each product is produced? b. Write in your own words how one determines the amount of excess reagent that remains unreacted after a reaction is complete. (What is the starting point for calculations? What conversion factors are used? What calculation is necessary to finish the determination?) Exercises A. Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example: SrHz(s) + 2 H200 → Sr(OH)2(s) + 2H2 (9) You wish to calculate the mass of hydrogen gas that can be prepared from 5.63 g of strontium hydride and 4.80 g of water. i.) If all of the strontium hydride reacts, how many moles of hydrogen gas will be produced? ii.) If all of the water reacts, how many moles of hydrogen gas will be produced?