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Stoichiometry

Learn how to solve limiting reactant problems and calculate the theoretical yield of a reaction using an example with potassium superoxide (KO2) and water (H2O).

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Stoichiometry

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  1. Stoichiometry Limiting Reagents

  2. Stoichiometry If the quantity of each reactant is given, you must determine which one is used up first. This is the Limiting Reagent The other reactant is not completely consumed in the reaction. It is in Excess.

  3. Stoichiometry Like many problems in chemistry, there are multiple ways of solving these problems. You can save yourself some work, by first identifying what you need to solve for and then choosing the appropriate method. Don’t worry if you choose the wrong method, you’ll get to the correct answer, it will just take you a few more steps.

  4. 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.

  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) First copy down the the BALANCED equation! Now place numerical the information below the compounds.

  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 0.10 mol Hide one Two starting amounts? Where do we start?

  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? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide Based on: KO2 2 mol H2O 0.15 mol KO2 = 0.075 mol H2O are needed 4 mol KO2 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol 0.10 mol We calculated that we need 0.075 moles of water, but we have 0.10 moles, thus water is in excess, thus our limiting reactant is KO2

  8. 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. KO2(s) • Determine the moles of oxygen produced 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol excess ? moles Based on: KO2 0.15 mol KO2 = mol O2 0.1125

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