1. Solution Stoichiometry In solution stoichiometry, we are given a concentration and a volume which we use to determine moles. n = C x V Then we use molar ratios to determine the unknown. Previous Unit: mass  moles  molar ratios  moles  mass mass  moles  molar ratios  moles  mass CURRENT Unit: C x V  moles  molar ratios  moles  Conc. C x V  moles  molar ratios  moles  Conc. In the past we used mass values given to us to determine moles of each reactant or product using molar ratios.

2. Finding Concentration or Volume of a Reactant  both concentration and volume are required to determine moles.  You will be given the concentration and volume of one reactant to the determine moles of the other reactant  Then use this mole value & molar ratios to i) determine the volume of the other reactant using the given concentration and molar ratios V = n / C ii) or the concentration of the other reactant using the given volume and molar ratios C = n / V

3. Example 1: A 10.0 mL sample of sulfuric acid reacts completely with 16.0 mL of 0.150 mol/L potassium hydroxide solution. Calculate the molar concentration of the sulfuric acid. 1 st – Balance the equation, convert mL to L 2 nd – Determine moles of one reactant C = n/V 3 rd - Use molar ratios to determine the moles needed of the other reactant 4 th - Use C = n/V to find the unknown value ( in this case concentration ) of the other reactant H 2 SO 4(aq) + 2 KOH (aq)  2 H 2 O (l) + K 2 SO 4 (aq) V = 10.0 mL V = 16.0 mL C = ? C = 0.150 mol/L

4. Example: A 10.0 mL sample of sulfuric acid reacts completely with 16.0 mL of 0.150 mol/L potassium hydroxide solution. Calculate the molar concentration of the sulfuric acid. 1 H 2 SO 4(aq) + 2 KOH (aq)  2 H 2 O (l) + K 2 SO 4(aq) V = 0.010 L V = 0.0160 L C = ? C = 0.150 mol/L n = C / V = 0.150 mol x 0.0160 L L = 0.00240 mol 1 : 2 x : 0.00240 x = 0.00240 / 2 = 0.00120

5. Example: A 10.0 mL sample of sulfuric acid reacts completely with 16.0 mL of 0.150 mol/L potassium hydroxide solution. Calculate the molar concentration of the sulfuric acid. 1 H 2 SO 4(aq) + 2 KOH (aq)  2 H 2 O (l) + K 2 SO 4(aq) V = 0.0100 L V = 0.0160 L C = ? C = 0.150 mol/L n = 0.00240 mol 1 : 2 x : 0.00240 x = 0.00120 (mol) C = n x V = 0.00120 mol x 0.0100 L = 0.120 mol / L

6. Example 2: A 0.200 mol/L sample of percarbonic acid reacts completely with 15.0 mL of 0.300 mol/L lithium hydroxide solution. Calculate the volume of the percarbonic acid. 1 H 2 CO 4(aq) + 2 LiOH (aq)  2 H 2 O (l) + Li 2 CO 4(aq) V = ? L V = 0.0150 L C = 0.200 mol/L C = 0.300 mol/L n = 0.0045 mol 1 : 2 x : 0.00450 x = 0.00225 (mol) V = n / C = 0.00225 mol = 0.0113 L or 11.3 mL 0.200 mol/L

7. Some reactions involve liquids and solids so we must determine moles from mass and use the C = n / V formula to determine the unknown quantity. Ex. 80.2 grams of Calcium reacted in a 250 mL of hydrochloric acid. What was the concentration of calcium chloride in solution at the end of the reaction? Ca (s) + 2 HCl (aq)  CaCl 2(aq) + H 2(g) 80.2 g V = 0.250 L V = 0.250 L n = m/Mm (note the total volume is = 80.2 g / 40.1 g/mol the same after the reaction) = 2.00 mol Ca : CaCl 2 1 : 1 C = n / V 2.00 : x = 2.00 mol / 0.250 L x = 2.00 mol = 8.00 mol/L of CaCl 2

8. HW pg 353 # 1-3 pg 358 # 9, 10 ( subtract 1 st ) Worksheet questions

9. Limiting Reactant Question When given the concentration and volume of both reactants or another way to determine moles then one must determine which reactant will run out first. Determine the amount of lead (II) iodide formed in grams when 100 mL of 2.50 mol/L of Pb(NO 3 ) 2 solution reacts with 200 mL of 3.50 mol/L KI solution. Pb(NO 3 ) 2(aq) + 2 KI (aq)  PbI 2(s) + 2 KNO 3(aq) 0.100 L 0.200 L m=? 2.50 mol/L 3.50 mol/L n = C x V n = C x V n = 0.250 mol n = 0.700 mol Which is the limiting reactant?

10. Determine the amount of lead (II) iodide formed in grams when 100 mL of 2.50 mol/L of Pb(NO 3 ) 2 solution reacts with 200 mL of 3.50 mol/L KI solution. Pb(NO 3 ) 2(aq) + 2 KI (aq)  PbI 2(s) + 2 KNO 3(aq) 0.100 L 0.200 L m=? 2.50 mol/L 3.50 mol/L n = C x V n = C x V n = 0.250 mol n = 0.700 mol Which is the limiting reactant? Pb(NO 3 ) 2 : KI 1 : 2 x : 0.700 x = 0.350 mol so Pb(NO 3 ) 2 is limiting

11. Determine the amount of lead (II) iodide formed in grams when 100 mL of 2.50 mol/L of Pb(NO 3 ) 2 solution reacts with 200 mL of 3.50 mol/L KI solution. Pb(NO 3 ) 2(aq) + 2 KI (aq)  PbI 2(s) + 2 KNO 3(aq) 0.100 L 0.200 L m=? 2.50 mol/L 3.50 mol/L m = n x Mm n = 0.250 mol n = 0.700 mol = 0.250 mol x 461 g/mol (limiting reactant) = 115 g Pb(NO 3 ) 2 : PbI 2 1 : 1 0.250 : x x = 0.250 mol

12. Determine the amount of lead (II) iodide formed in grams when 100 mL of 2.50 mol/L of Pb(NO 3 ) 2 solution reacts with 200 mL of 3.50 mol/L KI solution. Pb(NO 3 ) 2(aq) + 2 KI (aq)  PbI 2(s) + 2 KNO 3(aq) 0.100 L 0.200 L m=? 2.50 mol/L 3.50 mol/L m = n x Mm n = 0.250 mol n = 0.700 mol = 0.250 mol x 461 g/mol (limiting reactant) = 115 g Pb(NO 3 ) 2 : PbI 2 1 : 1 0.250 : x x = 0.250 mol

13. Determine the amount of solid calcium oxide formed in grams when 150 mL of 2.00 mol/L of Ca(NO 3 ) 2 solution reacts with 250 mL of 3.00 mol/L Na 2 O solution.