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From Masterton and Hurley, Chemistry: Principles and ... · From Masterton and Hurley, Chemistry: Principles and Reactions, 5th ed., Chap 4 Homework 1: 1. How would you prepare the

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Page 1: From Masterton and Hurley, Chemistry: Principles and ... · From Masterton and Hurley, Chemistry: Principles and Reactions, 5th ed., Chap 4 Homework 1: 1. How would you prepare the

From Masterton and Hurley, Chemistry: Principles and Reactions, 5th ed., Chap 4Homework 1:1. How would you prepare the following from the solid and pure watera. 0.400 L of 0.155 M Sr(OH)2

0.0400 L x 0.155 mol Sr(OH)2 x 121.63 g Sr(OH)2 = 1 L soln 1 mol Sr(OH)2

b. 1.75 L of 0.333 M (NH4)2CO3

3. You are asked to prepare a 0.8500 M solution of aluminum nitrate. You find that youonly have 50.00 g of the solid.a. What is the maximum volume of solution that you can prepare?

50.00 g Al(NO3)3 x 1 mol Al(NO3)3 x 1 L Al(NO3)3 solution = 0.242 L of solution 243.07 g Al(NO3)3 0.8500 mol Al(NO3)3

b. How many milliliters of this prepared solution are required to furnish 0.5000 mol ofaluminum nitrate to a reaction?

0.5000 mol Al(NO3)3 x 1 L soln = 0.588 L or 588 mL 0.8500 mol Al(NO3)3

c. If 2.500 L of the prepared solution is required, how much more aluminum nitrate wouldyou need?

2.500 L soln x 0.8500 mol Al(NO3)3 x 243.07 g Al(NO3)3 = 516.52 g Al(NO3)3 total needed 1 L soln 1 mol Al(NO3)3

516.52 – 50.00 g = 466.52 g additional Al(NO3)3 needed

d. Fifty milliliters of a 0.450 M solution of aluminum nitrate is needed. How would youprepare the required solution of the solution prepared in part a?

0.050 L soln d x 0.450 mol Al(NO3)3 x 1 L soln A = 26.5 mL of soln a are required 1 L soln d 0.85 mol Al(NO3)3 it is then diluted to 50 mL

11. a. Mg2+ (aq) + 2 OH- (aq) --> Mg(OH)2 (s) b. Ag+ (aq) + Cl- (aq) --> AgCl (s) and Ba2+ (aq) + SO4- (aq) --> BaSO4 (aq)

13. a. NR b. 2 Ag+ (aq) + CO32- (aq) --> Ag2CO3 (s) c. 2 Co3+ (aq) + 3 CO32- (aq) --> Co2(CO3)3 (s) d. 3 Ba2+ (aq) + 2 PO43- (aq) --> Ba3(PO4)2 (s) e. NR

Page 2: From Masterton and Hurley, Chemistry: Principles and ... · From Masterton and Hurley, Chemistry: Principles and Reactions, 5th ed., Chap 4 Homework 1: 1. How would you prepare the

17. What volume of 0.2500 M cobalt(III) sulfate is required to react completely witha. 25.00 mL of 0.0315 M Ca(OH)2

Co3+ (aq) + 3 OH- (aq) --> Co(OH)3 (s)

0.025 L Ca(OH)2 soln x 0.0315 mol Ca(OH)2 x 2 mol OH- x 1 mol Co3+ x 1 mol Co2(SO4)3 x 1 L Co2(SO4)3 = 1 L of soln 1 mol Ca(OH)2 3 mol OH- 2 mol Co3+ 0.25 mol Co2(SO4)3

= 0.00105 L of the Co2(SO4)3 soln

or

Co2(SO4)3 (aq) + 3 Ca(OH)2 (aq) --> 3 CaSO4 (aq) + 2 Co(OH)3 (s)

0.025 L Ca(OH)2 soln x 0.0315 mol Ca(OH)2 x 1 mol Co2(SO4)3 x 1 L Co2(SO4)3 = 1 L of soln 3 mol Ca(OH)2 0.25 mol Co2(SO4)3

= 0.00105 L of the Co2(SO4)3 soln

b. 5.00 g Na2CO3

3 Na2CO3 (aq) + Co2(SO4)3 (aq) --> Co2(CO3)3 (s) + 3 Na2SO4(aq)

5.00 g Na2CO3 x 1 mol Na2CO3 x 1 mol Co2(SO4)3 x 1 L Co2(SO4)3 = 0.0629 L Co2(SO4)3

105.99 g Na2CO3 3 mol Na2CO3 0.25 mol Co2(SO4)3

c. 12.5 mL of 0.1249 M K3PO4.

2 K3PO4 (aq) + Co2(SO4)3 --> 2 CoPO4 (s) + 3 K2SO4 (aq)

0.0125 L K3PO4 soln x 0.1249 mol K3PO4 x 1 mol Co2(SO4)3 x 1 L Co2(SO4)3 soln = 1 L K3PO4 soln 2 mol K3PO4 0.25 mol Co2(SO4)3

= 0.00312 L Co2(SO4)3 soln