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11.3 Limiting Reactants

11.3 Limiting Reactants. Limiting Reactant. A limiting reactant in a chemical reaction is the substance that Is used up first. Stops the reaction. Limits the amount of product that can form. Reacting Amounts. In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon.

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11.3 Limiting Reactants

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  1. 11.3 Limiting Reactants

  2. Limiting Reactant A limiting reactant in a chemical reaction is the substance that • Is used up first. • Stops the reaction. • Limits the amount of product that can form.

  3. Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item?

  4. Reacting Amounts Four table settings can be made. Initially Use Left over plates 5 4 1 forks 6 4 2 spoons 4 4 0 knives 7 4 3 The limiting item is the spoon.

  5. Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

  6. Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.

  7. Limiting Reactants When 4.00 mol H2 is mixed with 2.00 mol Cl2,how many moles of HCl can form? H2(g) + Cl(g)  2HCl (g) 4.00 mol 2.00 mol ??? mol • Calculate the moles of product from each reactant, H2 and Cl2. • The limiting reactant is the one that produces the smaller amount of product.

  8. Limiting Reactants Using Moles HCl from H2 4.00 mol H2 x 2 mol HCl = 8.00 mol HCl 1 mol H2(not possible) HCl from Cl2 2.00 mol Cl2 x 2 mol HCl = 4.00 mol HCl 1 mol Cl2 (smaller number) The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.

  9. Checking Calculations

  10. Limiting Reactants Using Mass If 4.80 mol Ca mixed with 2.00 mol N2, which is the limiting reactant? 3Ca(s) + N2(g)  Ca3N2(s) Moles of Ca3H2 from Ca 4.80 mol Ca x 1 mol Ca3N2 = 1.60 mol Ca3N2 3 mol Ca (Ca used up) Moles of Ca3H2 from N2 2.00 mol N2 x 1 mol Ca3N2 = 2.00 mol Ca3N2 1 mol N2 (not possible) All Ca is used up when 1.60 mol Ca3N2 forms. Thus, Ca is the limiting reactant.

  11. Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H2 and 24.0 g O2 react? 2H2(g) + O2(g) 2H2O(l)

  12. Limiting Reactants Using Mass Calculate the grams of H2 for each reactant. H2: 8.00 g H2 x 1 mol H2 x 2 mol H2O x 18.02 g H2O 2.016 g H2 2 mol H2 1 mol H2O = 71.5 g H2O (not possible) O2: 24.0 g O2 x 1 mol O2 x 2 mol H2O x 18.02 g H2O 32.00 g O2 1 mol O2 1 mol H2O = 27.0 g H2O (smaller) O2 is the limiting reactant.

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