Page 141 - Theory and Problems of BEGINNING CHEMISTRY

130 CHEMICAL EQUATIONS [CHAP. 8

(e) Copper(II) chloride plus hydrosulfuric acid yields copper(II) sulfide plus hydrochloric acid.
(f ) Barium hydroxide plus chloric acid yields barium chlorate plus water.
(g) Copper(II) sulfate plus water yields copper(II) sulfate pentahydrate.
(h) Copper(II) chloride plus sodium iodide yields copper(I) iodide plus iodine plus sodium chloride.
Ans. (a) PCl 5 + 4H 2 O −→ H 3 PO 4 + 5 HCl
(b) NaOH + H 2 SO 4 −→ NaHSO 4 + H 2 O
(c)2 C 2 H 6 + 7O 2 −→ 4CO 2 + 6H 2 O
(d)2 C 8 H 18 + 17 O 2 −→ 16 CO + 18 H 2 O
(e) CuCl 2 + H 2 S −→ CuS + 2 HCl
(f) Ba(OH) 2 + 2 HClO 3 −→ Ba(ClO 3 ) 2 + 2H 2 O
(g) CuSO 4 + 5H 2 O −→ CuSO 4 ·5H 2 O
(h) 2 CuCl 2 + 4 NaI −→ 2 CuI + I 2 + 4 NaCl

8.11. Balance the following equation: Cr + CrCl 3 −→ CrCl 2
Ans. Balance the Cl first, since the Cr appears in two reactants. Here, the Cr happens to be balanced automatically.

Cr + 2 CrCl 3 −→ 3 CrCl 2

8.12. Write balanced chemical equations for the following reactions: (a) Hydrogen chloride is produced by
the reaction of hydrogen and chlorine. (b) Hydrogen combines with chlorine to yield hydrogen chloride.
(c) Chlorine reacts with hydrogen to give hydrogen chloride.
Ans. (a), (b), and (c). H 2 + Cl 2 −→ 2 HCl

8.13. Write balanced chemical equations for the following reactions: (a) Hydrogen fluoride is produced by the
reaction of hydrochloric acid and sodium fluoride. (b) Hydrochloric acid combines with sodium fluoride
to yield hydrogen fluoride. (c) Sodium fluoride reacts with hydrochloric acid to give hydrofluoric acid.
Ans. (a), (b), and (c). HCl + NaF −→ NaCl + HF

8.14. Balance the following chemical equations:
(a)Na 2 CO 3 + HClO 3 −→ NaClO 3 + CO 2 + H 2 O
(b)Ca(HCO 3 ) 2 + HCl −→ CaCl 2 + CO 2 + H 2 O
(c) BiCl 3 + H 2 O −→ BiOCl + HCl
(d)H 2 S + O 2 −→ H 2 O + SO 2
(e)Cu 2 S + O 2 −→ Cu + SO 2
(f )NH 3 + O 2 −→ NO + H 2 O
(g)H 2 O 2 −→ H 2 O + O 2
Ans. (a)Na 2 CO 3 + 2 HClO 3 −→ 2 NaClO 3 + CO 2 + H 2 O
(b)Ca(HCO 3 ) 2 + 2 HCl −→ CaCl 2 + 2CO 2 + 2H 2 O
(c) BiCl 3 + H 2 O −→ BiOCl + 2 HCl
(d)2 H 2 S + 3O 2 −→ 2H 2 O + 2SO 2
(e)Cu 2 S + O 2 −→ 2Cu + SO 2
(f )4 NH 3 + 5O 2 −→ 4NO + 6H 2 O

(g)2 H 2 O 2 −→ 2H 2 O + O 2