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Stoichiometry overview Recall that in stoichiometry the mole ratio provides a necessary conversion factor: grams (x) moles (x) moles (y) grams (y) molar.

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Presentation on theme: "Stoichiometry overview Recall that in stoichiometry the mole ratio provides a necessary conversion factor: grams (x) moles (x) moles (y) grams (y) molar."— Presentation transcript:

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2 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 xmolar 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 xmol/L of y mole ratio from balanced equation Read pg Try Q 1-3a.

3 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? Pg. 353, Question 1 1 mol H 2 SO 4 2 mol NH 3 x # mol H 2 SO 4 = L NH mol H 2 SO 4 = 2.20 mol NH 3 L NH 3 x H 2 SO 4 (aq) + 2NH 3 (aq) (NH 4 ) 2 SO 4 (aq) Calculate mol H 2 SO 4, then mol/L = mol/ L mol/L = mol H 2 SO 4 / L = mol/L

4 Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of mol/L aluminum sulfate solution. Pg. 353, Question 2 3 mol Ca(OH) 2 1 mol Al 2 (SO 4 ) 3 x # L Ca(OH) 2 = L Al 2 (SO 4 ) 3 = L Ca(OH) mol Al 2 (SO 4 ) 3 L Al 2 (SO 4 ) 3 x Al 2 (SO 4 ) 3 (aq) + 3Ca(OH) 2 (aq) 2Al(OH) 3 (s) + 3CaSO 4 (s) L Ca(OH) mol Ca(OH) 2 x

5 A chemistry teacher wants 75.0 mL of mol/L iron(Ill) chloride solution to react completely with an excess of mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed? Pg. 353, Question 3 3 mol Na 2 CO 3 2 mol FeCl 3 x # L Na 2 CO 3 = L FeCl 3 = L Na 2 CO 3 = 90.0 mL Na 2 CO mol FeCl 3 L FeCl 3 x 2FeCl 3 (aq) + 3Na 2 CO 3 (aq) Fe 2 (CO 3 ) 3 (s) + 6NaCl(aq) L Na 2 CO mol Na 2 CO 3 x

6 Answers 1 mol H 2 SO 4 2 mol NH 3 x # mol H 2 SO 4 = L NH mol H 2 SO 4 = 2.20 mol NH 3 L NH 3 x 1. H 2 SO 4 (aq) + 2NH 3 (aq) (NH 4 ) 2 SO 4 (aq) Calculate mol H 2 SO 4, then mol/L = mol/ L mol/L = mol H 2 SO 4 / L = mol/L 3 mol Ca(OH) 2 1 mol Al 2 (SO 4 ) 3 x # L Ca(OH) 2 = L Al 2 (SO 4 ) 3 = L Ca(OH) mol Al 2 (SO 4 ) 3 L Al 2 (SO 4 ) 3 x 2. Al 2 (SO 4 ) 3 (aq) + 3Ca(OH) 2 (aq) 2Al(OH) 3 (s) + 3CaSO 4 (s) L Ca(OH) mol Ca(OH) 2 x

7 Answers 3 mol Na 2 CO 3 2 mol FeCl 3 x # L Na 2 CO 3 = L FeCl 3 = L Na 2 CO 3 = 90.0 mL Na 2 CO mol FeCl 3 L FeCl 3 x 3. 2FeCl 3 (aq) + 3Na 2 CO 3 (aq) Fe 2 (CO 3 ) 3 (s) + 6NaCl(aq) L Na 2 CO mol Na 2 CO 3 x

8 1.H 2 SO 4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H 2 SO 4 will be required to react completely with 75 mL of 0.50 mol/L NaOH? 2.How many moles of Fe(OH) 3 are produced when 85.0 L of iron(III) sulfate at a concentration of mol/L reacts with excess NaOH? 3.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

9 4.a) What volume of 0.20 mol/L AgNO 3 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? 5.What mass of precipitate should result when L of mol/L aluminum nitrate solution is mixed with L of 1.50 mol/L sodium hydroxide solution? Assignment

10 Answers 2 mol Fe(OH) 3 1 mol Fe 2 (SO 4 ) 3 x # mol Fe(OH) 3 = 85 L Fe 2 (SO 4 ) mol Fe 2 (SO 4 ) 3 L Fe 2 (SO 4 ) 3 x 1. H 2 SO 4 (aq) + 2NaOH(aq) 2H 2 O + Na 2 SO 4 (aq) = 102 mol 1 mol H 2 SO 4 2 mol NaOH x # L H 2 SO 4 = L NaOH = L = 9.4 mL 0.50 mol NaOH L NaOH x 2. Fe 2 (SO 4 ) 3 (aq) + 6NaOH(aq) 2Fe(OH) 3 (s) + 3Na 2 SO 4 (aq) L H 2 SO mol H 2 SO 4 x

11 3. 2NaOH (aq) + ZnCl 2 (aq) Zn(OH) 2 (s) + 2NaCl (aq) 1 mol Zn(OH) 2 2 mol NaOH x # g Zn(OH) 2 = L NaOH = 6.21 g 2.50 mol NaOH L NaOH x 4a. 3AgNO 3 (aq) + K 3 PO 4 (aq) Ag 3 PO 4 (s) + 3KNO 3 (aq) g Zn(OH) 2 1 mol Zn(OH) 2 x 3 mol AgNO 3 1 mol K 3 PO 4 x # L AgNO 3 = L K 3 PO 4 = L = 0.19 L 0.50 mol K 3 PO 4 L K 3 PO 4 x L AgNO mol AgNO 3 x 1 mol Ag 3 PO 4 1 mol K 3 PO 4 x # g Ag 3 PO 4 = L K 3 PO 4 = 5.2 g 0.50 mol K 3 PO 4 L K 3 PO 4 x g Ag 3 PO 4 1 mol Ag 3 PO 4 x 4b. 3AgNO 3 (aq) + K 3 PO 4 (aq) Ag 3 PO 4 (s) + 3KNO 3 (aq)

12 1 mol Al(OH) 3 1 mol Al(NO 3 ) 3 x # g Al(OH) 3 = L Al(NO 3 ) g Al(OH) 3 = g Al(OH) 3 1 mol Al(OH) 3 x mol Al(NO 3 ) 3 L Al(NO 3 ) 3 x 5. Al(NO 3 ) 3 (aq) + 3NaOH(aq) Al(OH) 3 (s) + 3NaNO 3 (aq) 1 mol Al(OH) 3 3 mol NaOH x # g Al(OH) 3 = L NaOH 9.36 g Al(OH) 3 = g Al(OH) 3 1 mol Al(OH) 3 x 1.50 mol NaOH L NaOH x For more lessons, visit


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