Solutions stoichiometry.

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Presentation transcript:

Solutions stoichiometry

Stoichiometry overview Recall that in stoichiometry the mole ratio provides a necessary conversion factor: grams (x)  moles (x)  moles (y)  grams (y) molar mass of x molar mass of y mole ratio from balanced equation We can do something similar with solutions: volume (x)  moles (x)  moles (y)  volume (y) mol/L of x mol/L of y mole ratio from balanced equation Read pg. 351-353. Try Q 1-3a.

Pg. 353, Question 1 Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used? H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Calculate mol H2SO4, then mol/L = mol/0.0500 L # mol H2SO4= 2.20 mol NH3 L NH3 x 1 mol H2SO4 2 mol NH3 x 0.0244 L NH3 0.02684 mol H2SO4 = mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L

Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) Pg. 353, Question 2 Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of 0.0250 mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of 0.125 mol/L aluminum sulfate solution. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) # L Ca(OH)2= 0.0250 L Al2(SO4)3 0.125 mol Al2(SO4)3 L Al2(SO4)3 x 3 mol Ca(OH)2 1 mol Al2(SO4)3 x L Ca(OH)2 0.0250 mol Ca(OH)2 x = 0.375 L Ca(OH)2

2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) Pg. 353, Question 3 A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with an excess of 0.250 mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed? 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) # L Na2CO3= 0.0750 L FeCl3 0.200 mol FeCl3 L FeCl3 x 3 mol Na2CO3 2 mol FeCl3 x L Na2CO3 0.250 mol Na2CO3 x = 0.0900 L Na2CO3 = 90.0 mL Na2CO3

2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) Answers 1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Calculate mol H2SO4, then mol/L = mol/0.0500 L # mol H2SO4= 2.20 mol NH3 L NH3 x 1 mol H2SO4 2 mol NH3 x 0.0244 L NH3 0.02684 mol H2SO4 = mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L 2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) # L Ca(OH)2= 0.0250 L Al2(SO4)3 0.125 mol Al2(SO4)3 L Al2(SO4)3 x 3 mol Ca(OH)2 1 mol Al2(SO4)3 x L Ca(OH)2 0.0250 mol Ca(OH)2 x = 0.375 L Ca(OH)2

3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) Answers 3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) # L Na2CO3= 0.0750 L FeCl3 0.200 mol FeCl3 L FeCl3 x 3 mol Na2CO3 2 mol FeCl3 x L Na2CO3 0.250 mol Na2CO3 x = 0.0900 L Na2CO3 = 90.0 mL Na2CO3

Assignment H2SO4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H2SO4 will be required to react completely with 75 mL of 0.50 mol/L NaOH? How many moles of Fe(OH)3 are produced when 85.0 L of iron(III) sulfate at a concentration of 0.600 mol/L reacts with excess NaOH? What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution.

Assignment a) What volume of 0.20 mol/L AgNO3 will be needed to react completely with 25.0 mL of 0.50 mol/L potassium phosphate? b) What mass of precipitate is produced from the above reaction? What mass of precipitate should result when 0.550 L of 0.500 mol/L aluminum nitrate solution is mixed with 0.240 L of 1.50 mol/L sodium hydroxide solution?

2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq) Answers 1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq) # L H2SO4= 0.075 L NaOH 0.50 mol NaOH L NaOH x 1 mol H2SO4 2 mol NaOH x L H2SO4 2.0 mol H2SO4 x = 0.009375 L = 9.4 mL 2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq) # mol Fe(OH)3= 0.600 mol Fe2(SO4)3 L Fe2(SO4)3 x 2 mol Fe(OH)3 1 mol Fe2(SO4)3 x 85 L Fe2(SO4)3 = 102 mol

# g Zn(OH)2= = 6.21 g 0.0500 L NaOH 2.50 mol NaOH L NaOH x 3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq) # g Zn(OH)2= = 6.21 g 0.0500 L NaOH 2.50 mol NaOH L NaOH x 1 mol Zn(OH)2 2 mol NaOH x 99.40 g Zn(OH)2 1 mol Zn(OH)2 x 4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # L AgNO3 = = 0.1875 L = 0.19 L 0.025 L K3PO4 0.50 mol K3PO4 L K3PO4 x 3 mol AgNO3 1 mol K3PO4 x L AgNO3 0.20 mol AgNO3 x 4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # g Ag3PO4= = 5.2 g 0.025 L K3PO4 0.50 mol K3PO4 L K3PO4 x 1 mol Ag3PO4 1 mol K3PO4 x 418.58 g Ag3PO4 1 mol Ag3PO4 x

5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq) # g Al(OH)3= 0.550 L Al(NO3)3 0.500 mol Al(NO3)3 L Al(NO3)3 x 1 mol Al(OH)3 1 mol Al(NO3)3 x 77.98 g Al(OH)3 1 mol Al(OH)3 x 21.4 g Al(OH)3 = # g Al(OH)3= 0.240 L NaOH 1.50 mol NaOH L NaOH x 1 mol Al(OH)3 3 mol NaOH x 77.98 g Al(OH)3 1 mol Al(OH)3 x 9.36 g Al(OH)3 = For more lessons, visit www.chalkbored.com