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Understanding Limiting and Excess Reactants in Chemistry Reactions

Learn about limiting reactants, excess reactants, and how to find the amount of product produced in chemical reactions through examples and step-by-step explanations.

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Understanding Limiting and Excess Reactants in Chemistry Reactions

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  1. Limiting Reactants

  2. Limiting vs. Excess • Limiting Reactant- The reactant in a chemical reaction that limits the amount of product that can be formed.  The reaction will stop when all of the limiting reactant is consumed. • Excess Reactant- The reactant in a chemical reaction that remains when the reaction stops. The excess reactant remains because there is nothingleft with which it can react.

  3. Finding the Limiting Reactant • Find the number of moles of one product for each reactant. • The reactant that produces the smallest amount of product is the limiting reactant. • involves doing a one or two step stoichiometry problem.

  4. Example Problem #1 • A 2.00 mol sample of ammonia is mixed with 4.00 mol of oxygen.  Which is the limiting reactant and which is in excess? 4 NH3+ 5 O2 4 NO + 6 H2O

  5. 4 NH3+ 5 O24 NO + 6 H2O NH3 is Limiting Reactant 4 mol NO 2.00 mol NH3 = 2.00 mol NO 4 mol NH3 O2 is Excess Reactant 4 mol NO 4.00 mol O2 = 3.20 mol NO 5 mol O2

  6. Finding the Amount of Product • Phase 1: Calculate the number of moles of one product for all reactants. • Phase 2: Use limiting reactant to calculate the amount of product(s) • involves doing a one or two step stoichiometry problem for each product you need to find

  7. Example Problem #2 • A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  What is the mass of nitrogen monoxide & water produced?

  8. Phase 1: Solve for moles of one product for ALLreactants • 4 NH3+ 5 O24 NO(g) + 6 H2O NH3 is Excess Reactant 4 mol NO 2.00 g NH3 1 mol NH3 = 0.117 mol NO 17.0 g NH3 4 mol NH3 O2 is Limiting Reactant 4 mol NO 4.00 g O2 1 mol O2 = 0.100 mol NO 32.0 g O2 5 mol O2

  9. If needed…Phase 2: Solve for mass of products Once you determine the limiting reactant, use that to solve any other problems. Finding mass of NO:From step 1 O2 was L.R., so start with the answer from O2 0.100 mol NO 30.0 g NO = 3.00 g NO produced 1 mol NO Finding mass of H2O:From step 1 O2 was L.R., so start with the answer from O2 0.100 mol NO 6 mol H2O 18.0 g H2O = 2.70 g H2O produced 4 mol NO 1 mol H2O

  10. Example Problem #3 • When 0.65 moles of oxygen reacts with 0.56 moles of potassium how many grams of potassium oxide will be produced? • 4 K + O22 K2O Phase 1: 0.65 mol O2 2 mol K2O = 1.3 mol K2O 1 mol O2 O2 is Excess Reactant 0.56 mol K 2 mol K2O = 0.28 mol K2O 4 mol K K is Limiting Reactant Phase 2: 0.28 mol K2O 94.2 g K2O = 26 g K2O produced 1 mol K2O

  11. Additional Examples • How much CO2 forms when 3.7 moles of C6H6 reacts with 28.4 moles of oxygen. • How much Al2O3 will form if 21.0 g of Al is reacted with 50.0 g of Fe2O3? 2C6H6 + 15O2→ 12CO2 + 6H2O 2Al + Fe2O3→ Al2O3 + 2Fe

  12. Additional Examples • If 15.3 g of silver nitrate reacts with 28.4 g of zinc chloride, what mass of silver chloride is produced? 2AgNO3 + ZnCl2→ 2AgCl + Zn(NO3)2

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