Stoichiometry overview Recall that in stoichiometry the mole

Pg. 353, Question 1 Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia.

Pg. 353, Question 2 Calcium hydroxide is sometimes used in water treatment plants to

Pg. 353, Question 3 A chemistry teacher wants 75. 0 m. L of 0.

Answers 1. H 2 SO 4(aq) + 2 NH 3(aq) (NH 4)2 SO 4(aq)

Answers 3. 2 Fe. Cl 3(aq) + 3 Na 2 CO 3(aq) Fe 2(CO

Assignment 1. H 2 SO 4 reacts with Na. OH, producing water and sodium

Assignment 4. a) What volume of 0. 20 mol/L Ag. NO 3 will be

Answers 1. H 2 SO 4(aq) + 2 Na. OH(aq) 2 H 2 O

3. 2 Na. OH(aq) + Zn. Cl 2(aq) Zn(OH)2(s) + # g Zn(OH)2= 2

5. Al(NO 3)3(aq) + 3 Na. OH(aq) Al(OH)3(s) + 3 Na. NO 3(aq) #g

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

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 m. L of a 2. 20 mol/L ammonia solution if 50. 0 m. L of sulfuric acid is used? H 2 SO 4(aq) + 2 NH 3(aq) (NH 4)2 SO 4(aq) Calculate mol H 2 SO 4, then mol/L = mol/0. 0500 L # mol H 2 SO 4= 2. 20 mol NH 31 mol H 2 SO 4 0. 02684 0. 0244 L NHx = mo x 3 L NH 3 2 mol NH 3 H 2 SO 4 mol/L = 0. 02684 mol H 2 SO 4 / 0. 0500 L =

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 m. L of 0. 125 mol/L aluminum sulfate solution. Al 2(SO 4)3(aq) + 3 Ca(OH)2(aq) 2 Al(OH)3(s) + 3 Ca. SO 4(s) # L Ca(OH)2= 0. 02500. 125 mol Al 2(SO 43)3 mol Ca(OH)2 L x x x L Ca(OH) L Al 2(SO 4)3 1 mol Al 2(SO 4)30. 0250 mol 2 Al 2(SO 4)3 Ca(OH)2 = 0. 375 L Ca(OH)2

Pg. 353, Question 3 A chemistry teacher wants 75. 0 m. L 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? 2 Fe. Cl 3(aq) + 3 Na 2 CO 3(aq) Fe 2(CO 3)3(s) + 6 Na. Cl(aq) # L Na 2 CO 3= 0. 0750 L 0. 200 mol Fe. Cl 33 mol Na 2 CO 3 L x x x Fe. Cl 3 Na 2 mol CO L Fe. Cl 3 2 mol Fe. Cl 3 0. 250 Na 23 CO 3 = 0. 0900 L Na 2 CO 3 = 90. 0 m. L Na 2 CO 3

Answers 1. H 2 SO 4(aq) + 2 NH 3(aq) (NH 4)2 SO 4(aq) Calculate mol H 2 SO 4, then mol/L = mol/0. 0500 L # mol H 2 SO 4= 2. 20 mol NH 31 mol H 2 SO 4 0. 02684 0. 0244 L NHx = mol x 3 L NH 3 2 mol NH 3 H 2 SO 4 mol/L = 0. 02684 mol H 2 SO 4 / 0. 0500 L = 0. 537 mol/L 2. Al 2(SO 4)3(aq) + 3 Ca(OH)2(aq) 2 Al(OH)3(s) + 3 Ca. SO 4(s) # L Ca(OH)2= 0. 02500. 125 mol Al 2(SO 43)3 mol Ca(OH)2 L x x x L Ca(OH) L Al 2(SO 4)3 1 mol Al 2(SO 4)30. 0250 mo 2 Al 2(SO 4)3 Ca(OH)2 = 0. 375 L Ca(OH)2

Answers 3. 2 Fe. Cl 3(aq) + 3 Na 2 CO 3(aq) Fe 2(CO 3)3(s) + 6 Na. Cl(aq) # L Na 2 CO 3= 0. 0750 L 0. 200 mol Fe. Cl 33 mol Na 2 CO 3 L x x x Fe. Cl 3 Na 2 mol CO L Fe. Cl 3 2 mol Fe. Cl 3 0. 250 Na 23 CO 3 = 0. 0900 L Na 2 CO 3 = 90. 0 m. L Na 2 CO 3

Assignment 1. H 2 SO 4 reacts with Na. OH, producing water and sodium sulfate. What volume of 2. 0 M H 2 SO 4 will be required to react completely with 75 m. L of 0. 50 mol/L Na. OH? 2. 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 Na. OH? 3. What mass of precipitate will be produced from the reaction of 50. 0 m. L of 2. 50 mol/L sodium hydroxide with an excess of zinc chloride solution.

Assignment 4. a) What volume of 0. 20 mol/L Ag. NO 3 will be needed to react completely with 25. 0 m. L 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 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?

Answers 1. H 2 SO 4(aq) + 2 Na. OH(aq) 2 H 2 O + Na 2 SO 4(aq) # L H 2 SO 4= mol Na. OH 0. 075 L 0. 50 1 mol H SO L H SO 2 4 x x x L Na. OH 2 mol Na. OH 2. 0 mol H 2 SO = 0. 009375 L = 9. 4 m. L 2. Fe 2(SO 4)3(aq) + 6 Na. OH(aq) 2 Fe(OH)3(s) + 3 Na SO (aq) 2 4 # mol Fe(OH) = 3 mol Fe 2(SO 4)3 2 mol Fe(OH)3 85 L Fe 2(SO 4)30. 600 x x L Fe 2(SO 4)3 1 mol Fe 2(SO 4)3 = 102 mol

3. 2 Na. OH(aq) + Zn. Cl 2(aq) Zn(OH)2(s) + # g Zn(OH)2= 2 Na. Cl(aq) = 6. 21 g mol Na. OH 1 mol Zn(OH) 99. 40 g 0. 05002. 50 2 x x x L Na. OH 2 mol Na. OH 1 mol Zn(OH) L Na. OH Zn(OH) 2 2 4 a. 3 Ag. NO 3(aq) + K 3 PO 4(aq) Ag 3 PO 4(s) + 3 KNO 3(aq) # L Ag. NO 3 = = 0. 1875 L = 0. 19 L 0. 025 0. 50 mol K 3 PO 34 mol Ag. NO 3 L Ag. NO 3 x x x L K 3 PO 4 1 mol K 3 PO 40. 20 mol Ag. NO 4 b. 3 Ag. NO 3(aq) + K 3 PO 4(aq) Ag 3 PO 4(s) + 3 KNO 3(aq) # g Ag 3 PO 4= = 5. 2 g mol K 3 PO 14 mol Ag 3 PO 4 418. 58 g 0. 025 0. 50 x x x L K 3 PO 4 1 mol Ag. Ag 3 PO 4 4 3 PO

5. Al(NO 3)3(aq) + 3 Na. OH(aq) Al(OH)3(s) + 3 Na. NO 3(aq) #g Al(OH)3=0. 500 mol 0. 550 x L L Al(NO 3)3 1 mol Al(OH)3 77. 98 g x x 1 mol 1 Al(OH) mol 3 Al(NO Al(OH) = 3)321. 4 g Al(OH) 3 3 #g Al(OH)1. 50 0. 240 mol Na. OH 1 mol Al(OH) 77. 98 g Al(OH) 3= 3 x x x L Na. OH 3 mol L Na. OH 1 mol Na. OH = 9. 36 Al(OH) g Al(OH) 3 3 For more lessons, visit www. chalkbored. com