1 / 11

STOICHIOMETRY TUTORIAL

STOICHIOMETRY TUTORIAL. Theoretical Yield and % Composition D. Lindsey. Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps.

meryl
Télécharger la présentation

STOICHIOMETRY TUTORIAL

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. STOICHIOMETRY TUTORIAL Theoretical Yield and % Composition D. Lindsey

  2. Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps. Get a pencil and paper, a periodic table and a calculator, and let’s get to work.

  3. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) First copy down the the BALANCED equation! Now place the numerical information below the compounds.

  4. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide one Two starting amounts? Where do we start?

  5. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide Based on: KO2 0.15 mol KO2 = mol O2 0.1125

  6. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles Hide 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 0.10 mol H2O Based on: H2O = mol O2 0.150

  7. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 It was limited by the amount of KO2. 0.10 mol H2O Based on: H2O = mol O2 0.150 H2O = excess (XS) reactant! What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?

  8. Theoretical yield vs.Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

  9. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide one 120.0 g KO2 Based on: KO2 = g O2 40.51

  10. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Question if only 35.2 g of O2 were recovered, what was the percent yield?

  11. If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H2O. 125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water. 84.79 g O2 31.83 = g XS H2O

More Related