Chapter 12 Homework Problems
36.
a)
Two formula units of KClO3 react to form 2 formula units of KCl and 3 molecules
of O2.
b)
Four molecules of NH3 react with 6 molecules of NO to form 5 molecules of N2
and 6 molecules of H2O.
c)
Four atoms of K react with 1 molecule of O2 to form 2 formula units of K2O.
37.
Just replace each “particle” word with mole in the above problem.
39.
A mole ratio is a ratio of coefficients from a balanced equation. It is used when
calculating the amount of an unknown reactant or product.
40.
a)
2.7 𝑚𝑜𝑙 𝐶 ×
b)
5.44 𝑚𝑜𝑙 𝑆𝑂2 ×
c)
0.246 𝑚𝑜𝑙 𝐶𝑆2 ×
d)
118 𝑚𝑜𝑙 𝐶𝑆2 ×
a)
3.60 × 102 𝑔 𝐶𝐻3 𝑂𝐻 ×
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
3.60 × 102 𝑔 𝐶𝐻3 𝑂𝐻 ×
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
41.
b)
1 𝑚𝑜𝑙 𝐶𝑆2
5 𝑚𝑜𝑙 𝐶
5 𝑚𝑜𝑙 𝐶
2 𝑚𝑜𝑙 𝑆𝑂2
4 𝑚𝑜𝑙 𝐶𝑂
2 𝑚𝑜𝑙 𝑆𝑂2
1 𝑚𝑜𝑙 𝐶𝑆2
4.00 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻 ×
2.85 𝑚𝑜𝑙 𝐶𝑂 ×
= 13.6 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶𝑆2
4.00 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻 ×
c)
= 0.54 𝑚𝑜𝑙 𝐶𝑆2
= 0.984 𝑚𝑜𝑙 𝐶𝑂
= 236 𝑚𝑜𝑙 𝑆𝑂2
32.0 𝑔 𝐶𝐻3 𝑂𝐻
32.0 𝑔 𝐶𝐻3 𝑂𝐻
2 𝑚𝑜𝑙 𝐻2
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
1 𝑚𝑜𝑙 𝐶𝑂
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
2 𝑚𝑜𝑙 𝐻2
1 𝑚𝑜𝑙 𝐶𝑂
×
×
×
2.0 𝑔 𝐻2
1 𝑚𝑜𝑙 𝐻2
×
×
1 𝑚𝑜𝑙 𝐶𝑂
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
2 𝑚𝑜𝑙 𝐻2
1 𝑚𝑜𝑙 𝐶𝐻3 𝑂𝐻
2.0 𝑔 𝐻2
1 𝑚𝑜𝑙 𝐻2
28.0 𝑔 𝐶𝑂
1 𝑚𝑜𝑙 𝐶𝑂
= 11.3 𝑚𝑜𝑙 𝐶𝑂
= 22.5 𝑚𝑜𝑙 𝐻2
= 16 𝑔 𝐻2
= 112 𝑔 𝐶𝑂
= 11 𝑔 𝐻2
42.
a)
66.6 𝑔 𝑁𝐻3 ×
b)
4.65 𝑔 𝐻𝐹 ×
c)
225 𝑔 𝐹2 ×
1 𝑚𝑜𝑙 𝑁𝐻3
17.0 𝑔 𝑁𝐻3
1 𝑚𝑜𝑙 𝐻𝐹
20.0 𝑔 𝐻𝐹
1 𝑚𝑜𝑙 𝐹2
38.0 𝑔 𝐹2
5 𝑚𝑜𝑙 𝐹2
×
×
2 𝑚𝑜𝑙 𝑁𝐻3
2 𝑚𝑜𝑙 𝑁𝐻3
6 𝑚𝑜𝑙 𝐻𝐹
1 𝑚𝑜𝑙 𝑁2 𝐹4
×
×
5 𝑚𝑜𝑙 𝐹2
1 𝑚𝑜𝑙 𝐹2
17.0 𝑔 𝑁𝐻3
×
×
38.0 𝑔 𝐹2
1 𝑚𝑜𝑙 𝑁𝐻3
104.0 𝑔 𝑁2 𝐹4
1 𝑚𝑜𝑙 𝑁2 𝐹4
= 372 𝑔 𝐹2
= 1.32 𝑔 𝑁𝐻3
= 123 𝑔 𝑁2 𝐹4
43.
Coefficients indicate mole ratios, particle ratios, or even volume (L) ratios.
44.
a)
32.9 𝑔 𝐿𝑖3 𝑁 ×
1 𝑚𝑜𝑙 𝐿𝑖3 𝑁
b)
32.9 𝑔 𝐿𝑖3 𝑁 ×
1 𝑚𝑜𝑙 𝐿𝑖3 𝑁
34.7 𝑔 𝐿𝑖3 𝑁
34.7 𝑔 𝐿𝑖3 𝑁
×
×
3 𝑚𝑜𝑙 𝐻2 𝑂
1 𝑚𝑜𝑙 𝐿𝑖3 𝑁
1 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝐿𝑖3 𝑁
×
×
18.0 𝑔 𝐻2 𝑂
1 𝑚𝑜𝑙 𝐻2 𝑂
= 51.2 𝑔 𝐻2 𝑂
6.02 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁𝐻3
=
5.71 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁𝐻3
c)
15.0 𝐿 𝑁𝐻3 ×
1 𝑚𝑜𝑙 𝑁𝐻3
22.4 𝐿 𝑁𝐻3
×
1 𝑚𝑜𝑙 𝐿𝑖3 𝑁
1 𝑚𝑜𝑙 𝑁𝐻3
×
34.7 𝑔 𝐿𝑖3 𝑁
1 𝑚𝑜𝑙 𝐿𝑖3 𝑁
= 23.2 𝑔 𝐿𝑖3 𝑁
45.
The limiting reagent determines how much product is formed during a chemical
reaction. The limiting reagent is completely used up, while the excess reagent will have
some left over.
47.
a)
To find the LR via Method 2, perform 2 calculations to determine how much
product (AlCl3) each reactant forms. Whichever one produces less reactant is the
LR.
3.0 𝑚𝑜𝑙 𝐴𝑙 ×
5.3 𝑚𝑜𝑙 𝐶𝑙2 ×
2 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3
2 𝑚𝑜𝑙 𝐴𝑙
= 3.0 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3
2 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3
3 𝑚𝑜𝑙 𝐶𝑙2
= 3.5 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3
b)
3.0 mol of AlCl3 will be formed
c)
3.0 𝑚𝑜𝑙 𝐴𝑙 ×
3 𝑚𝑜𝑙 𝐶𝑙2
2 𝑚𝑜𝑙 𝐴𝑙
Al is the LR
= 4.5 𝑚𝑜𝑙 𝐶𝑙2 𝑈𝑆𝐸𝐷!
5.3 𝑚𝑜𝑙 𝐶𝑙2 𝑔𝑖𝑣𝑒𝑛 − 4.5 𝑚𝑜𝑙 𝐶𝑙2 𝑢𝑠𝑒𝑑 = 0.8 𝑚𝑜𝑙 𝐶𝑙2 𝑒𝑥𝑐𝑒𝑠𝑠
48.
𝑎𝑐𝑡𝑢𝑎𝑙 (𝑙𝑎𝑏)𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 (𝑠𝑡𝑜𝑖𝑐ℎ.)𝑦𝑖𝑒𝑙𝑑
Remember, % 𝑦𝑖𝑒𝑙𝑑 =
× 100
Theoretical yield of Sb:
15.0 𝑔 𝑆𝑏2 𝑆3 ×
% 𝑦𝑖𝑒𝑙𝑑 =
50.
a)
9.84 𝑔
10.8 𝑔
1 𝑚𝑜𝑙 𝑆𝑏2 𝑆3
339.9 𝑔 𝑆𝑏2 𝑆3
2 𝑚𝑜𝑙 𝑆𝑏
×
1 𝑚𝑜𝑙 𝑆𝑏2 𝑆3
121.8 𝑔 𝑆𝑏
×
= 10.8 𝑔 𝑆𝑏
1 𝑚𝑜𝑙 𝑆𝑏
× 100 = 91.1 %
1.49 𝑔 𝐻𝑁𝑂3 ×
1 𝑚𝑜𝑙 𝐻𝑁𝑂3
63.0 𝑔 𝐻𝑁𝑂3
×
4 𝑚𝑜𝑙 𝑍𝑛
10 𝑚𝑜𝑙 𝐻𝑁𝑂3
6.02 × 1023 𝑎𝑡𝑜𝑚𝑠 𝑍𝑛
×
1 𝑚𝑜𝑙 𝑍𝑛
=
5.70 × 1021 𝑎𝑡𝑜𝑚𝑠 𝑍𝑛
b)
53.
29.1 𝑔 𝑁𝐻4 𝑁𝑂3 ×
1 𝑚𝑜𝑙 𝑁𝐻4 𝑁𝑂3
80.0 𝑔 𝑁𝐻4 𝑁𝑂3
×
4 𝑚𝑜𝑙 𝑍𝑛
1 𝑚𝑜𝑙 𝑁𝐻4 𝑁𝑂3
×
65.4 𝑔 𝑍𝑛
×
136.2 𝑔 𝐶𝑎𝑆𝑂4
1 𝑚𝑜𝑙 𝑍𝑛
= 95.2 𝑔 𝑍𝑛
If 100% yield:
5.24 𝑘𝑔 𝑆𝑂2 ×
1000 𝑔
1 𝑘𝑔
×
1 𝑚𝑜𝑙 𝑆𝑂2
64.1 𝑔 𝑆𝑂2
×
2 𝑚𝑜𝑙 𝐶𝑎𝑆𝑂4
2 𝑚𝑜𝑙 𝑆𝑂2
1 𝑚𝑜𝑙 𝐶𝑎𝑆𝑂4
×
1 𝑘𝑔
1000 𝑔
=
11.1 𝑘𝑔 𝐶𝑎𝑆𝑂4
For a 96.8% yield:
(11.1 kg) (0.968) = 10.7 kg CaSO4
64.
First convert 100 g of each reactant into moles. The convert moles of each reactant into
moles of the same product:
100.0 𝑔 𝑀𝑛𝑂2 ×
100.0 𝑔 𝐾𝑂𝐻 ×
100.0 𝑔 𝑂2
×
100.0 𝑔 𝐶𝑙2
×
1 𝑚𝑜𝑙 𝑀𝑛𝑂2
86.9 𝑔 𝑀𝑛𝑂2
1 𝑚𝑜𝑙 𝐾𝑂𝐻
56.1 𝑔 𝐾𝑂𝐻
1 𝑚𝑜𝑙 𝑂2
32.0 𝑔 𝑂2
1 𝑚𝑜𝑙 𝐶𝑙2
71.0 𝑔 𝐶𝑙2
= 1.15 𝑚𝑜𝑙 𝑀𝑛𝑂2 ×
= 1.78 𝑚𝑜𝑙 𝐾𝑂𝐻
×
2 𝑚𝑜𝑙 𝐻2 𝑂
2 𝑚𝑜𝑙 𝑀𝑛𝑂2
2 𝑚𝑜𝑙 𝐻2 𝑂
4 𝑚𝑜𝑙 𝐾𝑂𝐻
= 3.13 𝑚𝑜𝑙 𝑂2
×
2 𝑚𝑜𝑙 𝐻2 𝑂
= 1.41 𝑚𝑜𝑙 𝐶𝑙2
×
2 𝑚𝑜𝑙 𝐻2 𝑂
1 𝑚𝑜𝑙 𝑂2
1 𝑚𝑜𝑙 𝐶𝑙2
= 1.15 𝑚𝑜𝑙 𝐻2 𝑂
= 0.890 𝑚𝑜𝑙 𝐻2 𝑂
= 6.26 𝑚𝑜𝑙 𝐻2 𝑂
= 2.82 𝑚𝑜𝑙 𝐻2 𝑂
KOH produced
lowest amt of H2O
and is the LR.