Chemistry Notes: Reactions in Solutions
Download
1 / 5

Chemistry Notes: Reactions in Solutions - PowerPoint PPT Presentation


  • 85 Views
  • Uploaded on

Chemistry Notes: Reactions in Solutions Two solutions can be combined to generate a chemical reaction. Often, chemical reactions occur in aqueous * solutions, e.g. living systems.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Chemistry Notes: Reactions in Solutions' - althea


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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  • Chemistry Notes: Reactions in Solutions

  • Two solutions can be combined to generate a chemical reaction.

  • Often, chemical reactions occur in aqueous* solutions, e.g. living systems.

  • The concentration of each solution and the types of solutes involved must be taken into account. * Water = solvent

  • -------------------------------------------------------------------------------------

  • Example of a chemical reaction:

  • HCl + NaOH --> NaCl + H2O.

  • Let’s suppose this reaction occurs as a result of combining two solutions: HCl & NaOH.

  • Let: HCl solution = 1.0 M (= 1.0 mol HCl/L)

  • NaOH solution = 1.0 M (= 1.0 mol NaOH/L)

  • Combine 10 mL of each solution together. How many grams of NaCl are produced?

  • 1st: Determine how many moles of each reactant are used.

  • 10 mL of 1.0 M HCl = 10 mL x 1.0 mol = 0.01 mol HCl

  • 1000 L

  • 10 mL of 1.0 M HCl = 10 mL x 1.0 mol = 0.01 mol NaOH

  • 1000 L

Divide mL by 1000 to convert to L.


Next, determine moles of product produced (stoichiometry).

HCl + NaOH --> NaCl + H2O

0.01 + 0.01 --> 0.01 0.01 mol of NaCl produced.

mol molmol

Then, convert moles to grams.

(0.01 mol)(23+35)g/mol = 0.58 g NaCl produced

Molar mass NaCl


  • Example 2:

  • 15.0 mL of a 0.25 M solution of AgNO3 is combined with 10.0 mL of a 0.35 M solution of KBr in the following reaction:

  • AgNO3 + KBr ---> AgBr + KNO3 (balanced)

  • 1) What is the limiting reagent?

  • 2) How many grams of AgBr are produced?

  • Step 1: Determine the moles of each reactant used.

  • AgNO3 = (15 mL/1000)(0.25 mol/L) = 0.00375 mol AgNO3

  • KBr = (10 mL/1000)(0.35 mol/L) = 0.0035 mol KBr

  • Step 2: Determine limiting reagent.

  • AgNO3 + KBr ---> AgBr + KNO3

  • Molar ratios are all 1, so the limiting reagent will be determined by the smaller quantity of reactant, which is KBr @ 0.0035 mol.

  • Step 3: Determine the moles of product.

  • 0.0035 mol KBr ---> 0.0035 mol AgBr

  • Step 4: Calculate the grams of AgBr.

  • (0.0035 mol AgBr)(108+80)g/mol = 0.66 gAgBr produced


  • You do one.

  • 50 mL of a 5.00 M solution of Li2SO4 is combined with 45 mL of a 4.50 M solution of MgCl2, according to the following reaction:

  • Li2SO4 + MgCl2 ---> 2LiCl + MgSO4.

  •  How many grams of MgSO4 are produced?

  • Remember:

  • Step 1: Detemine the moles pf each reactant.

  • Step 2: Determine the limiting reagent.

  • Step 3: Determine the moles of the product.

  • Step 4: Calculate the grams of the product.

Molar mass AgBr


Solution

Step 1. (50 mL/1000)(5.00 M) = 0.2500 mol Li2SO4

(45 mL/1000)(4.50 M) = 0.2025 mol MgCl2

Step 2. Molar ratios for both reactants as well as the product are 1:1 so the smaller quantity of reactant (mol) = the quantity of product.

Smaller quantity of reactant = MgCl2 @ 0.2025 mol …

Step 3. … ---> 0.2025 mol MgSO4.

Step 4. (0.2025 mol MgSO4)(24+32+16x4) = 24.3 g MgSO4.


ad