Page 151 - Theory and Problems of BEGINNING CHEMISTRY

140 NET IONIC EQUATIONS [CHAP. 9

9.20. Balance the following net ionic equations:

(a)Ag + Fe −→ Fe 2+ + Ag (c) Cu 2+ + I −→ CuI + I 2
−
+
(b)Fe 3+ + I −→ Fe 2+ + I 2 (d)Cd + Cr 3+ −→ Cr 2+ + Cd 2+
−
Ans. In each part, the net charge as well as the number of each type of atom must balance.
(a)2 Ag + Fe −→ Fe 2+ + 2Ag (c)2 Cu 2+ + 4I −→ 2 CuI + I 2
−
+
−
(b)2 Fe 3+ + 2I −→ 2Fe 2+ + I 2 (d)Cd + 2Cr 3+ −→ 2Cr 2+ + Cd 2+
9.21. Try to write a complete equation corresponding to the unbalanced and the balanced net ionic equations of the prior
problem. What do you find?
Ans. You cannot write a complete equation for an unbalanced net ionic equation. [In part (a), for example, you
might have one acetate ion on the left and two on the right.] One complete equation for the balanced net
ionic equation might be
2 AgC H 3 O 2 + Fe −→ Fe(C 2 H 3 O 2 ) 2 + 2Ag
2
9.22. Would the following reaction yield 56 kJ of heat per mole of water formed, as the reactions in Example 9.4 do?
Explain.
HC 2 H 3 O 2 + NaOH −→ NaC 2 H 3 O 2 + H 2 O
Ans. No. Since HC 2 H 3 O 2 is a weak acid, there is a different net ionic equation and thus a different amount of
heat:
−
−
HC 2 H 3 O 2 + OH −→ C 2 H 3 O 2 + H 2 O
9.23. Per mole of water formed, how much heat is generated by the reaction of Example 9.5?
Ans. 56 kJ. It is the same reaction as that of Example 9.4.

9.24. Would 56 kJ per mole of water formed by generated by the following reaction? Compare your answer with that of
the prior problem.

Ba(OH) 2 (s) + 2 HCl −→ BaCl 2 + 2H 2 O
Ans. No. It is not represented by the same net ionic equation. Some heat is involved in dissolving the solid
Ba(OH) 2 .

9.25. Write a net ionic equation for each of the following overall equations:

(a) Ca(OH) 2 (s) + 2 HClO 3 −→ Ca(ClO 3 ) 2 + 2H 2 O (c) ZnS(s) + 2 HBr −→ H 2 S + ZnBr 2
(b) CuSO 4 (aq) + H 2 S −→ CuS(s) + H 2 SO 4
+
Ans. (a) Ca(OH) 2 (s) + 2H −→ Ca 2+ + 2H 2 O (c) ZnS(s) + 2H −→ H 2 S + Zn 2+
+
(b)Cu 2+ + H 2 S −→ CuS + 2H +
9.26. Write 12 more equations represented by the net ionic equation given in Example 9.4, using only the reactants used
in that example.
Ans. NaOH + HCl −→ NaCl + H 2 O NaOH + HBr −→ NaBr + H 2 O
RbOH + NHO 3 −→ RbNO 3 + H 2 O KOH + HI −→ KI + H 2 O
RbOH + HBr −→ RbBr + H 2 O LiOH + HCl −→ LiCl + H 2 O
NaOH + HI −→ NaI + H 2 O KOH + HNO 3 −→ KNO 3 + H 2 O
RbOH + HCl −→ RbCl + H 2 O KOH + HBr −→ KBr + H 2 O
LiOH + HNO 3 −→ LiNO 3 + H 2 O LiOH + HI −→ LiI + H 2 O