answers for exercises chapter 21 Petrucci, general chemistry...
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)
(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 .
(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
(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
Chapter 21: Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14
(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)
(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 ] xx x2 1.6 10 5 [CN ] 1.0 x 1.0 x Assume, x