1 / 12

Highland Science Department Limiting Reactants in Solution

Highland Science Department Limiting Reactants in Solution. Highland Science Department Limiting Reactants in Solution Limiting Reactant: the reactant that is used up first in a chemical reaction. Excess Reactant: the reactant that will still be left over when the reaction is complete.

yelena
Télécharger la présentation

Highland Science Department Limiting Reactants in Solution

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. Highland Science Department Limiting Reactants in Solution

  2. Highland Science Department Limiting Reactants in Solution Limiting Reactant: the reactant that is used up first in a chemical reaction. Excess Reactant: the reactant that will still be left over when the reaction is complete

  3. Highland Science Department Limiting Reactants in Solution Limiting Reactant: the reactant that is used up first in a chemical reaction. Excess Reactant: the reactant that will still be left over when the reaction is complete Finding Limiting Reactant: compare mole ratios to those needed in the balanced chemical reaction to see which one is present in lower numbers

  4. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide.

  5. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. G: Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(s) + 3CaSO4(aq)

  6. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. G: Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(s) + 3CaSO4(aq) volume = 20.0 mL concentration = 0.0150 mol/L Al2(SO4)3(aq) Al2(SO4)3(aq) = 0.0200 L volume = 30.0 mL concentration = 0.0185 mol/L Ca(OH)2(aq)Ca(OH)2(aq) = 0.0300 L M.M. Al(OH)3(s) = 78.0 g/mol

  7. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. U: mass of aluminum hydroxide

  8. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. U: mass of aluminum hydroxide S: 1. limiting reactant = volume x concentration ÷ lowest # moles n V M.R. M.M. 2. concentration  moles  moles  mass L.R. L.R. Al(OH)3 Al(OH)3

  9. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 1. moles = 0.0200 L = 3.0 x 10-4 mol Al2(SO4)3 moles = 0.0300 L = 5.55 x 10-4 mol Ca(OH)2

  10. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 1. moles = 0.0200 L = 3.0 x 10-4 mol Al2(SO4)3 3.0 x 10-4 mol = 1 equal to 1 mole moles = 0.0300 L = 5.55 x 10-4 mol Ca(OH)2 3.0 x 10-4 mol = 1.85 less than 3 moles

  11. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 1. moles = 0.0200 L = 3.0 x 10-4 mol Al2(SO4)3 3.0 x 10-4 mol = 1 equal to 1 mole Excess moles = 0.0300 L = 5.55 x 10-4 mol Ca(OH)2 3.0 x 10-4 mol = 1.85 less than 3 moles Limiting

  12. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 2. mass = 0.0300 L Ca(OH)2 Al(OH)3 = 0.0289 g

More Related