Page 143 - Theory and Problems of BEGINNING CHEMISTRY

132 CHEMICAL EQUATIONS [CHAP. 8

Ans. (a) Combination Cl 2 + 2 CrCl 2 −→ 2 CrCl 3
(b) Combination (or combustion) 2CO + O 2 −→ 2CO 2
heat
(c) Decomposition MgCO −→ CO 2 + MgO
3
(d) No reaction AlCl 3 + Cl 2 −→ NR
(e) Combustion 2 C 5 H 12 + 11 O 2 −→ 10 CO + 12 H 2 O
8.22. Which of the following is soluble in water, CuCl or CuCl 2 ?
Ans. CuCl 2 is soluble; CuCl is one of the four chlorides that are listed as insoluble (Table 8-2).

8.23. Complete and balance an equation for the reaction of excess HCl and Na 3 PO 4 .
Ans. 3 HCl + Na 3 PO 4 −→ H 3 PO 4 + 3 NaCl
The phosphoric acid produced is weak, that is, mostly covalent, and the formation of the H 3 PO 4 is the driving
force for this reaction. (HCl is one of the seven strong acids listed in Table 8-3.)

8.24. Is F 2 soluble in water?
Ans. No, it reacts with water, liberating oxygen:

2F 2 + 2H 2 O −→ 4HF + O 2

Supplementary Problems

8.25. What is the difference between dissolving and reacting?
Ans. Dissolving is a physical change, and no set ratio of substances is required.

8.26. How can you tell that the following are combination reactions rather than replacement or double-replacement
reactions.
CoCl 2 + Cl 2 −→ HgCl + Hg −→
2
CO + O 2 −→ BaO + CO 2 −→
Ans. The same element appears in both reactants.

8.27. State why the equation of Problem 8.2 is unusual.
Ans. It is a substitution reaction in which a nonmetal replaces a metal. Carbon, at high temperatures, can replace
relatively inactive metals from their oxides.

8.28. Complete and balance each of the following. If no reaction occurs, write “NR.”

(a)SO 2 + H 2 O −→ (l) C + O 2 (excess) −→
(b) BaO + H 2 O −→ (m)KOH + HNO 3 −→
(c) Al + O 2 −→ (n) HClO 3 + ZnO −→
(d)Na + S −→ (o) BaCl 2 + Na 2 SO 4 −→
(e) MgO + CO 2 −→ (p) AgNO + KCl −→
3
(f ) Al + Br 2 −→ (q) NaOH + H 2 SO 4 −→
(g)Ag + ZnCl 2 −→ (r) Ba(OH) 2 + HClO 2 −→
(h)Cl 2 + KBr −→ (s) C + O 2 (limited) −→
(i) BaSO 4 + NaCl −→ (t) C 5 H 10 + O 2 (excess) −→
electricity
(j) NaCl(l) −−−→ (u)C 5 H 10 + O 2 (limited) −→
(k) KCl + HNO 3 −→ (v) C 5 H 10 O + O 2 (excess) −→
(w)C 5 H 10 O + O 2 (limited) −→