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Chemistry Notes: Reactions in SolutionsPowerPoint Presentation

Chemistry Notes: Reactions in Solutions

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- 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

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.

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