Petrucci Chap 21 Answers

CHAPTER 21 CHEMISTRY OF THE MAIN-GROUP ELEMENTS I: GROUPS 1, 2, 13, AND 14 PRACTICE EXAMPLES 1A

(E) From Figure 21-2, the route from sodium chloride to sodium nitrate begins with electrolysis of NaCl(aq) to form NaOH(aq). 2 NaCl aq  + 2 H 2O l electrolysis   2 NaOH aq  + H 2 g  + Cl2 g  followed by addition of

NO2 g  to NaOH(aq). 2 NaOH  aq  + 3 NO 2  g   2 NaNO3  aq  + NO  g  + H 2 O(l)

1B

(E) From Figure 21-2, we see that the route from sodium chloride to sodium thiosulfate begins with the electrolysis of NaCl(aq) to produce NaOH(aq),   2 NaOH aq  + H 2 g  + Cl2 g  2 NaCl aq  + 2 H 2O l  electrolysis and continues through the reaction of SO2 g  with the NaOH(aq) in an acid-base reaction [ SO2 g  is an acid anhydride] to produce Na 2SO3  aq  :

2 NaOH  aq  + SO 2  g   Na 2SO3  aq  + H 2 O(l) and (3) concludes with

boil the addition of S to the boiling solution: Na 2SO3  aq  + S(s)  Na 2S2 O3  aq  .

2A

(M) The first reaction indicates that 0.1 mol of NaNO2 reacts with 0.3 mol of Na to yield 0.2 mol of compound X and 0.05 mol of N2. Since this reaction liberates nitrogen gas, it is very likely that compound X is an oxide of sodium. The two possibilities include Na2O and Na2O2. Na2O2 will not react with oxygen, but Na2O will. Therefore, compound X is most likely Na2O and compound Y is Na2O2. The balanced chemical equations for two processes are: 1 NaNO 2 (s)+3Na(s)  2Na 2O(s)+ N 2 (g) 2 1 Na 2O(s)+ O2 (g)  Na 2 O2 (s) 2

2B

(M) Since compound X is used in plaster of Paris it must contain calcium. Calcium reacts with carbon to form calcium carbide, according to the following chemical equation: Ca(s)+2C(s)  CaC2 (s)

Furthermore, compound X or calcium carbide reacts with nitrogen gas to form compound Y or calcium cyanamide: CaC2 (s)+N 2 (g)  CaCN 2 (s)+C(s) CN22- anion is isoelectronic with CO2 and it therefore contains 16 electrons. The structure of this anion is: N

C

N

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Chapter 21: Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14

3A

(M) The first two reactions, are those from Example 21-3, used to produce B2 O3 .

Na 2 B4 O7 10 H 2 O(s) + H 2SO 4 (l)   4 B  OH 3 (s) + Na 2SO 4 (s) + 5 H 2 O(l)

 2 B  OH 3 (s)  B2 O3 (s) + 3H 2 O(g)

The next reaction is conversion to BCl3 with heat, carbon, and chlorine.  2 B2 O3  s  + 3C  s  + 6 Cl2  g    4 BCl3  g  + 3CO 2  g 

LiAlH 4 is used as a reducing agent to produce diborane.

4 BCl3  g  + 3LiAlH 4 (s)  2 B2 H 6  g  + 3LiCl(s) + 3AlCl3 (s)

3B

(M) Na2B4O7.10 H2O(s) + H2SO4(aq)  4 B(OH)3(s) + Na2SO4(aq) + 5 H2O(l)   B2O3(s) + 3 H2O(l) 2 B(OH)3(s)   B2O3(s) + 3 CaF2(s) + 3 H2SO4(l)   2 BF3(g) + 3 CaSO4(s) + 3 H2O(g)

INTEGRATIVE EXAMPLE A

(M) NaCN and Al(NO3)3 dissociate in water according to the following equations: NaCN(aq)  Na + (aq)+CN - (aq) Al(NO 3 )3 (aq)  Al3+ (aq)+3NO-3 (aq) NaCN is a salt of strong base (NaOH) and weak acid (HCN) and its solution is therefore basic. We proceed by first determining the [OH]- concentration in a solution which is 1.0M in NaCN: CN +H 2O É HCN+OH 1.0 / / 1.0-x x x 14 Kw 1.0  10 Kb    1.6  10 5 10 K a (HCN ) 6.2  10 [HCN ][OH  ] xx x2    1.6  10 5  [CN ] 1.0  x 1.0  x Assume, x