1. 11.1 Section Assessment 7. How do you write a word equation?
To write a word equation, write the names of the reactants separated by plus signs followed by an arrow, followed by the names of the products separated by plus signs.
EXAMPLE: Ammonium hydroxide can be reacted with nitric acid to form ammonium nitrate and water.
The “word equation” for this would be:
Ammonium hydroxide + nitric acid  ammonium nitrate + water

2. 11.1 Section Assessment 8. How do you write a skeleton equation?
To write a skeleton equation, write the formulas of the reactants to the left of the arrow and the formulas of the products to the right of the arrow. (Don’t write any coefficients.)
EXAMPLE: Ammonium hydroxide can be reacted with nitric acid to form ammonium nitrate and water.
The “skeleton equation” for this would be:
NH4OH + HNO3  NH4NO3 + H2O

3. 11.1 Section Assessment
9. Describe the steps in writing a balanced chemical equation.
To write a balanced chemical equation, first write the skeleton equation for the reaction (see #8). Place whatever coefficients are needed to make the number of atoms of each element the same on the left of the arrow as on the right of the arrow.
EXAMPLE: Rusting is actually a pretty complicated process, but one of the net results is that iron metal combines with oxygen gas to form iron(III) oxide.
The “skeleton equation” for this would be:
Fe O2  Fe2O3 4 3 2
This equation isn’t “balanced”, however . . .
There is only one Fe on the left, but 2 on the right.
There are only 2 O’s on the left, but 3 on the right.
If we put the following coefficients in the equation, the number of atoms of each element become equal – “balanced”.

4. 11.1 Section Assessment
9. Describe the steps in writing a balanced chemical equation.
To write a balanced chemical equation, first write the skeleton equation for the reaction (see #8). Place whatever coefficients are needed to make the number of atoms of each element the same on the left of the arrow as on the right of the arrow.
EXAMPLE: Rusting is actually a pretty complicated process, but one of the net results is that iron metal combines with oxygen gas to form iron(III) oxide.
The “skeleton equation” for this would be:
Fe O2  Fe2O3 4 3 2
Fe: 4
Fe: 4 O: 6 O: 6

5. 11.1 Section Assessment
10.a. “Heating copper (II) sulfide in the presence of diatomic oxygen produces pure copper and sulfur dioxide gas”
translates into the following skeleton equation:
CuS + O Cu + SO2(g)
heat
10.b. “When heated, baking soda (sodium hydrogen carbonate) decomposes to form the products sodium carbonate, carbon dioxide, and water.”
translates into the following skeleton equation:
NaHCO Na2CO3 + CO2 + H2O
heat

6. 11.1 Section Assessment
11.a. “Iron metal and chlorine gas react to form solid iron(III) chloride,”
translates into the following balanced chemical equation: 2
Fe(s) + Cl2(g) FeCl3(s) 3 2 2
Fe:
Cl: 1 2
Fe:
Cl: 1 2 2 4 6 3 6

7. 11.1 Section Assessment
11.b. “Solid aluminum carbonate decomposes to form solid aluminum oxide and carbon dioxide gas,”
translates into the following balanced chemical equation:
Al2(CO3)3(s) Al2O3(s) CO2(g) 3
Al: 2
Al: 2 C: 3 C: 1 3 O: 9 O: 5 9

8. 11.1 Section Assessment
11.c. “Solid magnesium reacts with aqueous silver nitrate to form solid silver and aqueous magnesium nitrate,”
translates into the following balanced chemical equation:
Mg(s) AgNO3(aq) Ag(s) Mg(NO3)2(aq) 2 2
Mg: 1
Mg: 1
Ag: 1 2
Ag: 1 2 N: 1 2 N: 2 O: 3 6 O: 6

9. 11.1 Section Assessment 12.a. SO2 + O2 SO3 2 2 S: 1 2 S: 1 2 O: 4 6 O:
12.b Fe2O H Fe H2O 3 2 3
Fe: 2
Fe: 1 2 O: 3 O: 1 3 H: 2 6 H: 2 6

10. 11.1 Section Assessment 12.c. P + O2 P4O10 4 5 P: 1 4 P: 4 O: 2 10 O:
12.d Al N AlN 2 2
Al: 1 2
Al: 1 2 N: 2 N: 1 2

11. 11.2 Section Assessment
22. What are the five types of chemical reactions?
The five types of chemical reactions are . . .
. . . combination (also known as “synthesis”), decomposition, single-replacement (also known as “single displacement), double-replacement (also known as “double displacement), and combustion.

12. 11.2 Section Assessment
23. What are the keys to predicting the products of the five general types of reactions?
The key to predicting the products of the five general types of reactions is the number of elements and/or compounds reacting.

13. 11.2 Section Assessment
24. Classify each reaction and balance the equations.
a. C3H O2  CO H2O 2 9 6 6
= combustion
b. Al(OH)3  Al2O H2O 2 3
= decomposition
c. Li + O2  Li2O 4 2
= combination (aka “synthesis”)
d. Zn + AgNO3  Ag + Zn(NO3)2 2 2
= single replacement

14. 11.2 Section Assessment
25. Which of the five general types of reactions would most likely occur, given each set of reactants? What are the probably products?
a. an aqueous solution of two ionic compounds
double replacement; two different compounds
b. a single compound
decomposition; two or more elements and/or compounds
c. two elements
combination; a compound
d. oxygen and a comopund of carbon and hydrogen
combustion; carbon dioxide and water

15. 11.2 Section Assessment
26. Complete and balance an equation for each reaction
a. CaI2 + Hg(NO3)2  (HgI2 precipitates.)
CaI2 + Hg(NO3)2  HgI2 + Ca(NO3)2
b. Al + Cl2 
2Al + 3Cl2  2AlCl3
c. Ag + HCl 
No reaction. This might have been a single replacement reaction, but table 11.2 on page 333 indicates that Cu, Hg, and Ag will not replace H from acids, so H gets to stay with Cl and nobody changes.

16. 11.2 Section Assessment
26. Complete and balance an equation for each reaction
d. C2H2 + O2 
2C2H2 + 5O2  4CO2 + 2H2O
e. MgCl 
electricity
e. MgCl  Mg + Cl2

17. 11.2 Section Assessment
27. What are the three types of products that result from double-displacement reactions?
The three types of products that result from double displacement reactions are gas, precipitates, and molecular compounds.