Page 237 - Theory and Problems of BEGINNING CHEMISTRY
P. 237
226 SOLUTIONS [CHAP. 15
15.12. Calculate the mole fraction of the first component in each of the following solutions: (a)15.0gC 6 H 12 O 6
in 55.0 g H 2 O and (b)0.200 m solution of C 6 H 12 O 6 in water.
Ans. (a) 1 mol C 6 H 12 O 6
15.0gC H 12 O 6 = 0.0833 mol C 6 H 12 O 6
6
180gC H 12 O 6
6
1 mol H 2 O
55.0gH O = 3.06 mol H 2 O
2
18.0gH O
2
0.0833 mol
= = 0.0265
X C 6 H 12 O 6
3.06 mol + 0.0833 mol
(b) In 1.00 kg H 2 O there are 0.200 mol C 6 H 12 O 6 and
1 mol H 2 O
1000 g H O = 55.6 mol H 2 O
2
18.0gH O
2
The mole fraction is
0.200 mol
= = 0.00358
X C 6 H 12 O 6
55.8 mol total
EQUIVALENTS
15.13. An equivalent is defined as the amount of a substance that reacts with or produces 1 mol of what?
Ans. In a redox reaction, electrons. In an acid-base reaction, hydrogen ions or hydroxide ions.
15.14. What is the number of equivalents per mole of HCl?
Ans. Assuming complete neutralization (or even oxidation of the Cl to Cl 2 ), we can see that 1 mol HCl will react
−
with 1 mol NaOH (or 1 mol e ); thus, 1 equiv = 1 mol. (In its reactions to produce oxyanions of chlorine
−
−
(ClO , ClO 2 , etc.), there is more than 1 equivalent per mole.)
−
15.15. What numbers of equivalents per mole are possible for H 3 PO 4 in its acid-base reactions?
Ans. 1, 2, or 3, depending on how complete its reaction with base is.
15.16. How many equivalents of NaOH react with the 82.0 g of H 2 SO 3 in part a of Example 15.9?
−
Ans. 1 equiv. (One OH can react with 1 mol H ; 1 equiv of H 2 SO 3 is involved.)
+
15.17. How many equivalents are there per mole of the first reactant in each of the following equations?
(a)Zn −→ Zn 2+ + 2e − (f ) NaHCO 3 + HCl −→ NaCl + CO 2 + H 2 O
(b)H 2 SO 4 + NaOH −→ NaHSO 4 + H 2 O (g) NaH 2 PO 4 + 2 NaOH −→ Na 3 PO 4 + 2H 2 O
(c) 2 HCl + Ba(OH) 2 −→ BaCl 2 + 2H 2 O (h)2 H 2 SO 4 + Cu −→ CuSO 4 + SO 2 + 2H 2 O
− − + 2+ + 4H 2 O (i) NaHCO 3 + NaOH −→ Na 2 CO 3 + H 2 O
(d) MnO 4 + 5 e + 8H −→ Mn
+
(e)H 2 SO 4 + 2H + Cu −→
Cu 2+ + SO 2 + 2H 2 O
Ans. (a)2 (b)1 (c)1 (d)5 (e)2 (f)1 (g)2 (h) 2 [same as (e) for the one H 2 SO 4 molecule that is
reduced] (i)1
NORMALITY
15.18. If a bottle is labeled 3.109 N H 3 PO 4 , which reaction should be assumed for the definition of its normality?
Ans. H 3 PO 4 + 3 NaOH −→ Na 3 PO 4 + 3H 2 O (complete neutralization)
15.19. Explain why normality is the number of milliequivalents per milliliter.