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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 112<br />
- <strong>Chapter</strong> 9 -<br />
44. Composition of a mixture of and KCl.<br />
In the mixture only KCl reacts with AgNO 3 .<br />
KCl(aq) + AgNO 3 (aq) ¡ AgCl(s) + KNO 3 (aq)<br />
74.55 g KCl<br />
14.33 g AgCl2a b = 2.25 g KCl in the mixture<br />
143.7 g AgCl<br />
KNO 3<br />
- 112 -<br />
10.00 g mixture - 2.25 g KCl = 7.75 g KNO 3<br />
2.25 g KCl<br />
a b11002 = 22.5% KCl<br />
10.00 g mixture<br />
a 7.75 g KNO 3<br />
10.00 g mixture b11002 = 77.5% KNO 3<br />
45. The balanced equation is Zn + 2 HCl ¡ ZnCl 2 + H 2<br />
180.0 g Zn - 35 g Zn = 145 g Zn reacted with HCl<br />
(a)<br />
(b)<br />
(c)<br />
1145 g Zn2a 1 mol<br />
65.39 g ba1 mol H 2<br />
1 mol Zn b a 2.016 g b = 4.47 g H<br />
mol<br />
2 produced<br />
1145 g Zn2a 1 mol mol HCl<br />
ba2<br />
65.39 g 1 mol Zn ba36.46 g b = 162 g HCl reacted<br />
mol<br />
1180.0 g Zn2a 1 mol mol HCl<br />
ba2<br />
65.39 g 1 mol Zn ba36.46 g b = 201 g HCl reacts<br />
mol<br />
201 g - 162 g = 39 g more HCl needed to react wih the 180.0 g Zn<br />
46.<br />
Fe(s) + CuSO 4 (aq)<br />
2.0 mol 3.0 mol<br />
¡<br />
Cu(s)<br />
FeSO 4 (aq)<br />
(a) 2.0 mol Fe react with 2.0 mol CuSO 4 to yield 2.0 mol Cu and 2.0 mol FeSO 4 .<br />
1.0 mol CuSO 4 is unreacted. At the completion of the reaction, there will be<br />
2.0 mol Cu, 2.0 mol FeSO 4 , and 1.0 mol CuSO 4 .<br />
(b) Determine which reactant is limiting and then calculate the g FeSO 4 produced from<br />
that reactant.<br />
120.0 g Fe2a 1 mol mol Cu<br />
ba1<br />
55.85 g 1 mol Fe ba63.55 g b = 22.8 g Cu<br />
mol<br />
140.0 g CuSO 4 2a 1 mol 1 mol Cu<br />
ba ba 63.55 g b = 15.9 g Cu<br />
159.6 g 1 mol CuSO 4 mol<br />
Since CuSO 4 produces less Cu, it is the limiting reactant. Determine the mass of<br />
FeSO 4 produced from 40.0 g CuSO 4 .<br />
+
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