Transcript 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.

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.
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
87.91 g FeS
= 0.284 mol FeS
Step 2) 0.284 mol FeS 2 mol HCl
1 mol FeS
Step 3) 0.569 mol HCl
1 L Solution
2.5 mol HCl
= 0.569 mol HCl
1000 mL
1L
= 228 mL HCl
solution
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
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. mL KBr
1L
1.5 mol KBr = 0.675 mol KBr
1000 mL 1 L solution
Step 2) 0.675 mol KBr 1 mol Cl2
2 mol KBr
Step 3) 0.338 mol Cl2
70.90 g Cl2
1 mol Cl2
= 0.338 mol Cl2
= 23.9 g Cl2
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)
Dilutions
Stock solution - solutions with a high concentration, of commonly
used aqueous solutions
Dilute – decrease the concentration of a solution by adding more
solvent
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.
12.0 mol
= .729 L stock sol.
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
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?