Chapter 9

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 116
- Chapter 9 -
a 6.00 g NaHCO 3
b11002 = 60.0% NaHCO
10.00 g
3
a 4.00 g Na 2CO 3
b11002 = 40.0% Na
10.00 g
2 CO 3
54. The balanced equation is 2 KClO 3 ¡ 2 KCl + 3 O 2
12.82 g mixture - 9.45 g residue = 3.37 g O 2 lost by heating
Because the O 2 lost came only from KClO 3 , we can use it to calculate the amount of
KClO 3 in the mixture.
The conversion is: g O 2 ¡ mol O 2 ¡ mol KClO 3 ¡ g KClO 3
13.37 g O 2 2a 1 mol
32.00 g ba2 mol KClO 3
ba 122.6 g b = 8.61 g KClO
3 mol O 2 mol
3 in the mixture
a 8.61 g KClO 3
12.82 g sample b11002 = 67.2% KClO 3
55. The balanced equation is
Al(OH) 3 (s) + 3 HCl(aq) ¡ AlCl 3 (aq) + 3 H 2 O(l)
The conversion is: L HCl ¡ g HCl ¡ mol HCl ¡ mol Al(OH) 3 : g Al(OH) 3
a 2.5 L
day
g HCl
ba3.0 ba 1 mol
L
36.46 g b ¢ 1 mol Al1OH2 3
3 mol HCl
≤ a 78.00 g b = 5.3 g Al1OH2
mol
3 >day
Now calculate the number of 400. mg tablets that can be made from 5.3 g Al(OH) 3
¢ 5.3 g Al(OH) 3 1000 mg
≤ a ba 1 tablet
day
g 400. mg b = 13 tablets>day
56.
4 P + 5 O 2 ¡ P 4 O 10
P 4 O 10 + 6 H 2 O ¡ 4 H 3 PO 4
In the first reaction:
120.0 g P2a 1 mol b = 0.646 mol P
30.97 g
130.0 g O 2 2a 1 mol
32.00 g b = 0.938 mol O 2
0.646 mol P 3.44 mol P
This is a ratio of
=
0.938 mol O 2 5.00 mol O 2
Therefore, P is the limiting reactant and the P 4 O 10 produced is:
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