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Assignment 5.07: Solution Stoichiometry

Assignment 5.07: Solution Stoichiometry. Solution Concentration (Review of 5.06). Molarity = moles of solute/Liters of solution 25.2 g of NaCl dissolved in 500. mL of Water. Percent by Mass = mass of solute/mass of solution 25.2 g of NaCl dissolved in 500. mL of Water.

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Assignment 5.07: Solution Stoichiometry

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  1. Assignment 5.07: Solution Stoichiometry

  2. Solution Concentration (Review of 5.06) • Molarity = moles of solute/Liters of solution 25.2 g of NaCl dissolved in 500. mL of Water. • Percent by Mass = mass of solute/mass of solution 25.2 g of NaCl dissolved in 500. mL of Water.

  3. Solution Stoichiometry (Sample) How many milliliters of 2.5 M Hydrochloric acid solution would be needed to react with 25.0 g of Iron (II) sulfide? FeS + 2HCl  H2S + FeCl2 Step 1) 25.0 g FeS 1 mol FeS = 0.284 mol FeS 87.91 g FeS Step 2) 0.284 mol FeS 2 mol HCl = 0.569 mol HCl 1 mol FeS Step 3) 0.569 mol HCl 1 L Solution 1000 mL = 228 mLHCl 2.5 mol HCl 1 L solution

  4. Solution Stoichiometry (Practice) How many milliliters of 3.5 M Hydrochloric acid solution would be needed to react completely with 32.0 g of Magnesium metal? Mg + 2HCl  MgCl2 + H2

  5. Solution Stoichiometry (Sample) How many grams of chlorine gas are needed to react with 450. mL of a 1.5 M potassium bromide solution? Cl2 + 2KBr → 2KCl + Br2 Step 1) 450. mLKBr 1 L 1.5 mol KBr = 0.675 mol KBr 1000 mL 1 L solution Step 2) 0.675 mol KBr 1 mol Cl2 = 0.338 mol Cl2 2 mol KBr Step 3) 0.338 mol Cl2 70.90 g Cl2 = 23.9 g Cl2 1 mol Cl2

  6. Solution Stoichiometry (Practice) How many grams of aluminum are needed to react with 745 mL of a 2.5 M iron (II) nitrate solution? 2Al (s) + 3Fe(NO3)2 (aq) → 3Fe (s) + 2Al(NO3)2 (aq)

  7. Dilutions Stock solution - solutions with a high concentration, of commonly used aqueous solutions Dilute – decrease the concentration of a solution by adding more solvent

  8. Dilutions (Sample) How would a student make 3.5 liters of a 2.50 M solution of acetic acid from a 12.0 M stock solution? Step 1) 3.5 L 2.50 mol = 8.75 mol Acetic Acid 1 L dilute sol. Step2) 8.75 mol 1 L stock sol. = .729 L stock sol. 12.0 mol Step 3) 3.5 liters - .729 liters = 2.77 liters of water added  Measure .729 L of the 12.0 M stock solution, then add 2.77 L of water to make 3.5 L of 2.5 M solution

  9. Dilutions (Practice) How would a student make 100. mL of a 2.0 M solution of hydrochloric acid from a 12.0 M stock solution? How would a student make 100. mL of a 1.5 M solution from 18.0 M sulfuric acid?

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