Name the following acids

Naming Acids Worksheet Key Block_____
Name the following acids
1. HCL
2. HClO4
3. HIO3
4. HI
5. H2SO4
6. H2S
7. HCN
8. H2CO3
9. HBrO4
10. HBrO3
11. HC2H3O2
12. H3PO3
13. H3P
14. H2CrO4
15. H2CrO2
16. H2Cr
hydrochloric acid
perchloric acid
iodic acid
hydroiodic acid
sulfuric acid
hydrosulfuric acid
hydrocyanic acid
carbonic acid
perbromic acid
bromic acid
acetic acid
phosphorous acid
hydrophosphoric acid
chromic acid
hypochromous acid
hydrochromic acid
Give formulas for the following acids
17. hydrofluoric acid
18. nitric acid
19. nitrous acid
20. hydroxic acid
21. periodic acid
22. percarbonic acid
23. hypobromous acid
24. bromous acid
25. permanganic acid
26. manganic acid
27. iodous acid
28. sulfurous acid
29. sulfuric acid
30. perchloric acid
31. chlorous acid
32. acetic acid
33. phosphoric acid
34. phosphorous acid
HF
HNO3
HNO2
H2 O
HIO4
H2CO4
HBrO
HBrO2
HMnO4
HMnO3
HIO2
H2SO3
H2SO4
HClO4
HClO2
HC2H3O2
H3PO4
H3PO3
Name_______________________
5.6B
Solutions for 5.6A Extra Practice Questions:
Applications of Stoichiometry
1. In a chemical analysis to test the purity of a bottle of sodium bromide, a solution containing
1.17 g of sodium bromide was reacted with an excess of dimercury(I) acetate solution. The dry
precipitate had a mass of 2.73 g. Calculate the percent yield for the precipitate and comment
on the purity of sodium bromide.
2 NaBr(aq) Hg2(CH3COO)2(aq) → Hg2Br2(s) 2 NaCH3COO(aq)
1.17 g
m
102.89 g/mol
560.98 g/mol
1 mol
nNaBr 1.17 g 0.0114 mol
102.89 g
1
nHg Br 0.0114 mol 0.005 69 mol
2
2
2
560.98 g
mHg Br 0.005 69 mol 3.19 g
2
2
1 mol
1 mol NaBr
1 mol Hg2Br2
560.98 g Hg2Br2
or mHg Br 1.17 g NaBr 2
2
102.89 g NaBr
2 mol NaBr
1 mol Hg2Br2
3.19 g
2.73 g
% yield 100 85.6%
3.19 g
The purity of the sodium bromide is relatively poor and it is likely a technical grade not a
reagent grade (see Figure 7.8, page 261).
2. A solution containing 2.56 g of aluminum nitrate is mixed with a solution containing 1.02 g
of ammonium sulfide. Determine the unreacted mass of the excess reagent and the mass of
precipitate formed.
2 Al(NO3)3(aq) 3(NH4)2S(aq) → Al2S3(s) 6 NH4NO3(aq)
2.56 g
1.02 g
m
213.01 g/mol
68.16 g/mol
150.14 g/mol
1 mol
nAl(NO ) 2.56 g 0.0120 mol
33
213.01 g
1 mol
n(NH ) S 1.02 g 0.0150 mol
42
68.16 g
If Al(NO3)3 is the limiting reagent, the amount of (NH4)2S required is
3
n(NH ) S 0.0120 mol 0.0180 mol
42
2
Insufficient (NH4)2S is present. Therefore Al(NO3)3 is the excess reagent and (NH4)2S is
the limiting reagent.
2
nAl(NO ) 0.0150 mol 0.009 98 mol
(required amount)
33
3
nAl(NO ) 0.0120 mol 0.009 98 mol 0.0020 mol
(excess amount)
33
213.01 g
mAl(NO ) 0.0020 mol 0.43 g excess
(unreacted mass)
33
1 mol
1
nAl S 0.0150 mol 0.004 99 mol
2 3
3
150.14 g
mAl S 0.004 99 mol 0.749 g
2 3
1 mol
1 mol Al2S3
150.14 g Al2S3
or mAl S 0.0150 mol (NH4)2S 2 3
3 mol (NH4)2S
1 mol Al2S3
0.749 g
2-26
Unit 2 Lab and Study Masters
Copyright © 2002 Nelson Thomson Learning
5.6B
Solutions for 5.6A Extra Practice Questions:
Applications of Stoichiometry
1. In a chemical analysis to test the purity of a bottle of sodium bromide, a solution containing
1.17 g of sodium bromide was reacted with an excess of dimercury(I) acetate solution. The dry
precipitate had a mass of 2.73 g. Calculate the percent yield for the precipitate and comment
on the purity of sodium bromide.
2 NaBr(aq) Hg2(CH3COO)2(aq) → Hg2Br2(s) 2 NaCH3COO(aq)
1.17 g
m
102.89 g/mol
560.98 g/mol
1 mol
nNaBr 1.17 g 0.0114 mol
102.89 g
1
nHg Br 0.0114 mol 0.005 69 mol
2
2
2
560.98 g
mHg Br 0.005 69 mol 3.19 g
2
2
1 mol
1 mol NaBr
1 mol Hg2Br2
560.98 g Hg2Br2
or mHg Br 1.17 g NaBr 2
2
102.89 g NaBr
2 mol NaBr
1 mol Hg2Br2
3.19 g
2.73 g
% yield 100 85.6%
3.19 g
The purity of the sodium bromide is relatively poor and it is likely a technical grade not a
reagent grade (see Figure 7.8, page 261).
2. A solution containing 2.56 g of aluminum nitrate is mixed with a solution containing 1.02 g
of ammonium sulfide. Determine the unreacted mass of the excess reagent and the mass of
precipitate formed.
2 Al(NO3)3(aq) 3(NH4)2S(aq) → Al2S3(s) 6 NH4NO3(aq)
2.56 g
1.02 g
m
213.01 g/mol
68.16 g/mol
150.14 g/mol
1 mol
nAl(NO ) 2.56 g 0.0120 mol
33
213.01 g
1 mol
n(NH ) S 1.02 g 0.0150 mol
42
68.16 g
If Al(NO3)3 is the limiting reagent, the amount of (NH4)2S required is
3
n(NH ) S 0.0120 mol 0.0180 mol
42
2
Insufficient (NH4)2S is present. Therefore Al(NO3)3 is the excess reagent and (NH4)2S is
the limiting reagent.
2
nAl(NO ) 0.0150 mol 0.009 98 mol
(required amount)
33
3
nAl(NO ) 0.0120 mol 0.009 98 mol 0.0020 mol
(excess amount)
33
213.01 g
mAl(NO ) 0.0020 mol 0.43 g excess
(unreacted mass)
33
1 mol
1
nAl S 0.0150 mol 0.004 99 mol
2 3
3
150.14 g
mAl S 0.004 99 mol 0.749 g
2 3
1 mol
1 mol Al2S3
150.14 g Al2S3
or mAl S 0.0150 mol (NH4)2S 2 3
3 mol (NH4)2S
1 mol Al2S3
0.749 g
2-26
Unit 2 Lab and Study Masters
Copyright © 2002 Nelson Thomson Learning
7.3B
Solutions for 7.3A Extra Practice Questions: Net Ionic
Equations
Write nonionic, complete ionic, and net ionic equations to represent the following reactions.
1. A zinc strip is dipped into an aqueous solution of silver nitrate.
Zn(s) 2 AgNO3(aq) → 2 Ag(s) Zn(NO3)2(aq)
Zn(s) 2 Ag(aq) 2 NO3(aq) → 2 Ag(s) Zn2(aq) 2 NO3(aq)
Zn(s) 2 Ag(aq) → 2 Ag(s) Zn2(aq)
2. Solutions of lead(II) nitrate and potassium chloride are mixed.
Pb(NO3)2(aq) 2 KCl(aq) → 2 KNO3(aq) PbCl2(s)
Pb2(aq) 2 NO3(aq) 2 K(aq) 2 Cl(aq) → 2 K(aq) 2 NO3(aq) PbCl2(s)
Pb2(aq) 2 Cl(aq) → PbCl2(s)
3. Barium hydroxide is used to neutralize sulfuric acid.
Ba(OH)2(aq) H2SO4(aq) → 2 HOH(l) BaSO4(s)
Ba2(aq) 2 OH(aq) 2 H(aq) SO42(aq) → 2 HOH(l) BaSO4(s)
4. Chlorine gas is bubbled through an aqueous solution of sodium iodide.
2 NaI(aq) Cl2(g) → 2 NaCl(aq) I2(s)
2 Na(aq) 2 I(aq) Cl2(g) → 2 Na(aq) 2 Cl(aq) I2(s)
2 I(aq) Cl2(g) → 2 Cl(aq) I2(s)
5. Aqueous solutions of aluminum sulfate and calcium hydroxide are mixed.
Al2(SO4)3(aq) 3 Ca(OH)2(aq) → 3 CaSO4(s) 2 Al(OH)3(s)
2 Al3(aq) 3 SO42(aq) 3 Ca2(aq) 6 OH(aq) → 3 CaSO4(s) 2 Al(OH)3(s)
6. An iron nail is dropped into a beaker of sulfuric acid.
2 Fe(s) 3 H2SO4(aq) → Fe2(SO4)3(aq) 3 H2(g)
2 Fe(s) 6 H(aq) 3 SO42(aq) → 2 Fe3(aq) 3 SO42(aq) 3 H2(g)
2 Fe(s) 6 H(aq) → 2 Fe3(aq) 3 H2(g)
3-34
Unit 3 Lab and Study Masters
Copyright © 2002 Nelson Thomson Learning