Chapter 9

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 108
- Chapter 9 -
Now calculate the percent
in the impure sample.
33. No. There are not enough screwdrivers, wrenches or pliers. 2400 screwdrivers,
3600 wrenches and 1200 pliers are needed for 600 tool sets.
34. A subscript is used to indicate the number of atoms in a formula. It cannot be changed
without changing the identity of the substance. Coefficients are used only to balance atoms in
chemical equations. They may be changed as needed to achieve a balanced equation.
35. Consider the reaction A ¡ 2B and assume that you have 1 gram of A. This does not
guarantee that you will produce 1 gram of B because A and B have different molar
masses. One gram of A does not contain the same number of molecules as 1 gram of B.
However, 1 mole of A does have the same number of molecules as one mole of B.
(Remember, 1 mole = 6.022 * 10 23 molecules always.) If you determine the number of
moles in one gram of A and multiply by 2 to get the number of moles of B then from
that you can determine the grams of B using its molar mass. Equations are written in
terms of moles not grams.
36.
(a)
CaC 2
- 108 -
¢ 34.6 g CaC 2
44.5 g sample ≤11002 = 77.8% CaC 2 in the impure sample
4 KO 2 + 2 H 2 O + 4 CO 2 ¡ 4 KHCO 3 + 3 O 2
Á
¢ 0.85 g CO 2
≤ a 1 mol
min 44.01 g b ¢ 4 mol KO 2
≤ = 0.019 mol KO 2
4 mol CO 2 min
¢ 0.019 mol KO 2
b110.0 min2 = 0.19 mol KO
min
2
(b)
37. (a)
The conversion is:
g CO 2
min ¡ mol CO 2
min
¡ mol O 2
min
¢ 0.85 g CO 2
≤ a 1 mol
min 44.01 g b ¢ 3 mol O 2
≤ a 32.00 g
4 mol CO 2 mol
1750 g C 6 H 12 O 6 2a 1 mol
180.2 g b ¢ 2 mol C 2H 5 OH
≤ a 46.07 g b = 380 g C
1 mol C 6 H 12 O 6 mol
2 H 5 OH
ba
¡ g O 2
min ¡ g O 2
hr
60.0 min
b = 28 g O 2
1.0 hr hr
1750 g C 6 H 12 O 6 2a 1 mol
180.2 g b ¢ 2 mol CO 2
≤ a 44.01 g b = 370 g CO
1 mol C 6 H 12 O 6 mol
2
Alternate Solution: 750 g C 2 H 6 O 6 - 380 g
C 2 H 5 OH = 370 g CO 2 by the conservation of mass method.
(b)
1380 g C 2 H 5 OH2a 1 mL
0.79 g b = 480 mL C 2H 5 OH