Chemistry Homework Solutions

Chemistry/General Chemistry

Formulas/ equations

(See attached file for full problem description)

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1. Fe²+ and Sb³ ˉ
2. Mg²+ and As³ ˉ
3. Al³+ and O² ˉ
4. Co³+ and I ˉ
5. Sr² + and P³ ˉ
6. Ca²+ and C4+
7. Fe³+ and As³ ˉ
8. Na + and S² ˉ

Write the name or formula for these compounds:
1. tin (II) acetate
2. bromous acid
3. copper(II) permanganate
4. CI4
5. SiBr4

Balance the following equations
a. Na2SO3 (aq) + S8 (s) → Na2S2O3
b. C4H11O 2 (l) + O 2 (g)→
c. C12H22O11 (l) + O 2 (g) →

Write the molecular equation for the following reactions ( show physical states of both reactants and products)
a. potassium hydroxide and copper (II) sulfate
b. ammonium sulfide and iron(III) chloride
c. barium nitrate and sodium carbonate
d. sodium chloride and silver nitrate

Write the net ionic equation for the following reactions show physical states for both reactants and products.
a. potassium carbonate and iron (III) nitrate
b. ammonium bromide and lead (IV) nitrate
c. sodium hydroxide and gold (III) chloride
d. lithium sulfide and mercury (I) nitrate

  • chem 1.doc
chem 1.doc

Fe²+ and Sb³ ˉ

Mg²+ and As³ ˉ

Al³+ and O² ˉ

Co³+ and I ˉ

Sr² + and P³ ˉ

Ca²+ and C4+

Fe³+ and As³ ˉ

Na + and S² ˉ

Write the name or formula for these compounds:

tin (II) acetate

bromous acid

copper(II) permanganate

CI4

SiBr4

Balance the following equations

Na2SO3 (aq) + S8 (s) → Na2S2O3

C4H11O 2 (l) + O 2 (g)→

C12H22O11 (l) + O 2 (g) →

Write the molecular equation for the following reactions ( show physical
states of both reactants and products)

potassium hydroxide and copper (II) sulfate

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摧䚢±ഀ sulfide and iron(III) chloride

barium nitrate and sodium carbonate

sodium chloride and silver nitrate

Write the net ionic equation for the following reactions show physical
states for both reactants and products.

potassium carbonate and iron (III) nitrate

ammonium bromide and lead (IV) nitrate

sodium hydroxide and gold (III) chloride

lithium sulfide and mercury (I) nitrate


Naming Compounds, Balancing Equations, & Chemical Formulae

(see the attached Word document for a nicely formatted version of this.)

Stephen Allen
OTA 104330

===================================================

To do these questions, you need to make sure that the ionic compound formed is neutral. Therefore, the easy way to do this is to place the number next to the charge of one species as the subscript of the other species. For example, for the first one, we take the 2 next to the 2+ of Fe and put it as the subscript of Sb, making Sb2. Likewise; we take the 3 next to the 3- of Sb and put it as the subscript to Fe, making Fe3. Then put the two together making sure the cation (positively charged species) goes first. Do the same thing to each one:

1. Fe²+ and Sb³ ˉ  Fe3Sb2
2. Mg²+ and As³ ˉ  Mg3As2
3. Al³+ and O² ˉ  Al2O3
4. Co³+ and I ˉ  CoI3
5. Sr² + and P³ ˉ  Sr3P2
6. Ca²+ and C4+  Something is wrong here. You don't have any anion. Please check the question.
7. Fe³+ and As³ ˉ  FeAs
8. Na + and S² ˉ  Na2S

Write the name or formula for these compounds:
1. tin (II) acetate  Sn(C2H3O2)2
a. Each Sn (tin) cation has a +2 charge. Since acetate (C2H3O2) has a -1 charge, we need two of them to get the whole compound neutral.

2. bromous acid  HBrO2
a. HBrO is hypobromous acid

3. copper (II) permanganate  Cu(MnO4)2
a. Copper is +2 and permanganate is MnO4-. Therefore, you need two permanganates to each copper to get a neutral compound.

4. CI4  Carbon tetraiodide
a. For binary compounds that contain two nonmetals, name the first one, and then the second one with an "-ide" ending. Of course, you need a prefix in front of the second species to let everyone know how many of them there are.

5. SiBr4  Silicon tetrabromide

Balance the following equations
a. 8 Na2SO3 (aq) + S8 (s) → 8 Na2S2O3

b. 2 C4H11O 2 (l) + 11 O 2 (g) → 8 CO2 + 10 H2O (To do this, you first must know that when you react oxygen with an organic compound, you're probably having a combustion reaction which always produces carbon dioxide and water. Then you balance the carbons first, then the hydrogens, and then finally the oxygens. If you have a fraction with the oxygens as you would here, it being 11/2, then multiply everything by two to get final numbers.)

c. Here's how to do this one step by step:
C12H22O11 (l) + O 2 (g) → CO2 + H2O
C12H22O11 (l) + O 2 (g) → 12 CO2 + H2O
C12H22O11 (l) + O 2 (g) → 12 CO2 + 11 H2O
C12H22O11 (l) + 12 O 2 (g) → 12 CO2 + 11 H2O

Write the molecular equation for the following reactions (show physical states of both reactants and products)
a. potassium hydroxide and copper (II) sulfate
a. KOH + CuSO4  K2SO4 + Cu(OH)2
b. Now we balance the equation:
c. 2 KOH + CuSO4  K2SO4 + Cu(OH)2

b. ammonium sulfide and iron (III) chloride
a. (NH4)2S + FeCl3  NH4Cl + Fe2S3
b. Now we balance it.
c. 3 (NH4)2S + 2 FeCl3  6 NH4Cl + Fe2S3

c. barium nitrate and sodium carbonate
a. Ba(NO3)2 + Na2CO3  BaCO3 + NaNO3
b. Now we balance it:
c. Ba(NO3)2 + Na2CO3  BaCO3 + 2 NaNO3

d. sodium chloride and silver nitrate
a. NaCl + AgNO3  NaNO3 + AgCl

(I'll let you put in the physical states. It's not that hard. Just remember your rules of solubility. Consider most everything as "(aq)" or aqueous unless the species formed is insoluble. When a species is insoluble, right it as "(s)" meaning "solid.")

Write the net ionic equation for the following reactions show physical states for both reactants and products.
a. potassium carbonate and iron (III) nitrate
a. First, write out the chemical species like this, making sure to keep each compound neutral:
b. K2CO3 + Fe(NO3)3  KNO3 + Fe2(CO3)3
c. Then, balance the equation:
d. 3 K2CO3 + 2 Fe(NO3)3  6 KNO3 + Fe2(CO3)3
e. Next, put in the physical states of each:
f. 3 K2CO3 (aq) + 2 Fe(NO3)3 (aq)  6 KNO3 (aq) + Fe2(CO3)3 (s)
g. Eliminate "spectator" ions. The only chemical species really doing anything are the iron and carbonate species:
h. 3 CO32- (aq) + 2 Fe3+ (aq)  Fe2(CO3)3 (s)
i. We're done!

b. ammonium bromide and lead (IV) nitrate
a. NH4Br + Pb(NO3)4  NH4NO3 + PbBr4
b. 4 NH4Br + Pb(NO3)4  4 NH4NO3 + PbBr4
c. 4 NH4Br (aq) + Pb(NO3)4 (aq)  4 NH4NO3 (aq) + PbBr4 (s)
d. 4 Br- (aq) + Pb4+ (aq)  PbBr4 (s)

c. sodium hydroxide and gold (III) chloride
a. NaOH + AuCl3  NaCl + Au(OH)3
b. 3 NaOH + AuCl3  3 NaCl + Au(OH)3
c. 3 NaOH (aq) + AuCl3 (aq)  3 NaCl (aq) + Au(OH)3 (s)
d. 3 OH- (aq) + Au3+ (aq)  Au(OH)3 (s)

d. lithium sulfide and mercury (I) nitrate
a. You should be able to do this one by yourself now. Just follow the steps I've shown you already. You can do it!

  • chem1_BrainMass.doc
chem 1.doc

Fe²+ and Sb³ ˉ

Mg²+ and As³ ˉ

Al³+ and O² ˉ

Co³+ and I ˉ

Sr² + and P³ ˉ

Ca²+ and C4+

Fe³+ and As³ ˉ

Na + and S² ˉ

Write the name or formula for these compounds:

tin (II) acetate

bromous acid

copper(II) permanganate

CI4

SiBr4

Balance the following equations

Na2SO3 (aq) + S8 (s) → Na2S2O3

C4H11O 2 (l) + O 2 (g)→

C12H22O11 (l) + O 2 (g) →

Write the molecular equation for the following reactions ( show physical
states of both reactants and products)

potassium hydroxide and copper (II) sulfate

Ñ€

摧䚢±ഀ sulfide and iron(III) chloride

barium nitrate and sodium carbonate

sodium chloride and silver nitrate

Write the net ionic equation for the following reactions show physical
states for both reactants and products.

potassium carbonate and iron (III) nitrate

ammonium bromide and lead (IV) nitrate

sodium hydroxide and gold (III) chloride

lithium sulfide and mercury (I) nitrate

chem1_BrainMass.doc

To do these questions, you need to make sure that the ionic compound
formed is neutral. Therefore, the easy way to do this is to place the
number next to the charge of one species as the subscript of the other
species. For example, for the first one, we take the 2 next to the 2+ of
Fe and put it as the subscript of Sb, making Sb2. Likewise; we take the
3 next to the 3- of Sb and put it as the subscript to Fe, making Fe3.
Then put the two together making sure the cation (positively charged
species) goes first. Do the same thing to each one:

Fe²+ and Sb³ ˉ ( Fe3Sb2

Mg²+ and As³ ˉ ( Mg3As2

Al³+ and O² ˉ ( Al2O3

Co³+ and I ˉ ( CoI3

Sr² + and P³ ˉ ( Sr3P2

Ca²+ and C4+ ( Something is wrong here. You don’t have any anion.
Please check the question.

Fe³+ and As³ ˉ ( FeAs

Na + and S² ˉ ( Na2S

Write the name or formula for these compounds:

tin (II) acetate ( Sn(C2H3O2)2

Each Sn (tin) cation has a +2 charge. Since acetate (C2H3O2) has a -1
charge, we need two of them to get the whole compound neutral.

bromous acid ( HBrO2

HBrO is hypobromous acid

copper (II) permanganate ( Cu(MnO4)2

Copper is +2 and permanganate is MnO4-. Therefore, you need two
permanganates to each copper to get a neutral compound.

CI4 ( Carbon tetraiodide

For binary compounds that contain two nonmetals, name the first one, and
then the second one with an “-ide” ending. Of course, you need a
prefix in front of the second species to let everyone know how many of
them there are.

SiBr4 ( Silicon tetrabromide

Balance the following equations

8 Na2SO3 (aq) + S8 (s) → 8 Na2S2O3

2 C4H11O 2 (l) + 11 O 2 (g) → 8 CO2 + 10 H2O (To do this, you first
must know that when you react oxygen with an organic compound, you’re
probably having a combustion reaction which always produces carbon
dioxide and water. Then you balance the carbons first, then the
hydrogens, and then finally the oxygens. If you have a fraction with the
oxygens as you would here, it being 11/2, then multiply everything by
two to get final numbers.)

→ CO2 + H2O

C12H22O11 (l) + O 2 (g) → 12 CO2 + H2O

C12H22O11 (l) + O 2 (g) → 12 CO2 + 11 H2O

C12H22O11 (l) + 12 O 2 (g) → 12 CO2 + 11 H2O

Write the molecular equation for the following reactions (show physical
states of both reactants and products)

potassium hydroxide and copper (II) sulfate

KOH + CuSO4 ( K2SO4 + Cu(OH)2

Now we balance the equation:

2 KOH + CuSO4 ( K2SO4 + Cu(OH)2

ammonium sulfide and iron (III) chloride

(NH4)2S + FeCl3 ( NH4Cl + Fe2S3

Now we balance it.

3 (NH4)2S + 2 FeCl3 ( 6 NH4Cl + Fe2S3

barium nitrate and sodium carbonate

Ba(NO3)2 + Na2CO3 ( BaCO3 + NaNO3

Now we balance it:

Ba(NO3)2 + Na2CO3 ( BaCO3 + 2 NaNO3

sodium chloride and silver nitrate

NaCl + AgNO3 ( NaNO3 + AgCl

(I’ll let you put in the physical states. It’s not that hard. Just
remember your rules of solubility. Consider most everything as
“(aq)” or aqueous unless the species formed is insoluble. When a
species is insoluble, right it as “(s)” meaning “solid.”)

Write the net ionic equation for the following reactions show physical
states for both reactants and products.

potassium carbonate and iron (III) nitrate

First, write out the chemical species like this, making sure to keep
each compound neutral:

K2CO3 + Fe(NO3)3 ( KNO3 + Fe2(CO3)3

Then, balance the equation:



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摧Ⴖ-᐀ + 2 Fe(NO3)3 ( 6 KNO3 + Fe2(CO3)3

Next, put in the physical states of each:

3 K2CO3 (aq) + 2 Fe(NO3)3 (aq) ( 6 KNO3 (aq) + Fe2(CO3)3 (s)

Eliminate “spectator” ions. The only chemical species really doing
anything are the iron and carbonate species:

3 CO32- (aq) + 2 Fe3+ (aq) ( Fe2(CO3)3 (s)

We’re done!

ammonium bromide and lead (IV) nitrate

NH4Br + Pb(NO3)4 ( NH4NO3 + PbBr4

4 NH4Br + Pb(NO3)4 ( 4 NH4NO3 + PbBr4

4 NH4Br (aq) + Pb(NO3)4 (aq) ( 4 NH4NO3 (aq) + PbBr4 (s)

4 Br- (aq) + Pb4+ (aq) ( PbBr4 (s)

sodium hydroxide and gold (III) chloride

NaOH + AuCl3 ( NaCl + Au(OH)3

3 NaOH + AuCl3 ( 3 NaCl + Au(OH)3

3 NaOH (aq) + AuCl3 (aq) ( 3 NaCl (aq) + Au(OH)3 (s)

3 OH- (aq) + Au3+ (aq) ( Au(OH)3 (s)

lithium sulfide and mercury (I) nitrate

You should be able to do this one by yourself now. Just follow the steps
I’ve shown you already. You can do it!