Chemistry 105 Worksheet Chapter 6
1.
Balance the following equations.
+
10 Na (g) +
2 KNO3 (s)  1 K2O (s)+
2 NaHCO3 (s) 
2 C4H10(g)
2 H2O (l)

2 Mg (s)
1 Na2O(s)
+13 O2(g)
1 Li2CO3 (aq) +

+
2 Mg(OH)2 (aq)
+ 1
5 Na2O (s)
+
1 N2 (g)
2 CO2 (g)+
1 H2O (g)
8 CO2(g)
2 HCl (aq)  1 CO2 (g) +
+
1 H2O (l) +
H2(g)
10 H2O(g)
2 LiCl (aq)
2. For each atom in the above equations, tell me what the oxidation number on that
atom is. Which of the reactions is a redox equation? In those redox equations,
what are the oxidizing and reducing reagents?
eqn 1 Mg: 0, H: +1, O: -2 →
Mg:+2, O: -2, H in OH: +1, H in H2: 0
this is a redox eqn, Mg is reducing agent, and the H that goes to hydrogen gas is
oxidizing agent.
eqn 2. Na: 0, K: +1, N: +5, O: -2
→
K: +1, O: -2, Na: +1, O: -2, N: 0
also redox, Na is reducing agent, N is oxidizing agent
eqn 3. Na: +1, H: +1, C: +4, O: -2 → Na: +1, all O:-2, C+4, H +1
not a redox
eqn 4. done in class
eqn 5. Li +1, C +4, O -2, H +1, Cl -1
→ same again, no redox
3. Ammonia is one of the largest products of the chemical industry today. It is used
mainly for fertilizer production and is made via what is called the Haber process.
Basically, hydrogen and air are combined in a reaction vessel. The nitrogen in air
combines with the hydrogen to make ammonia. Because there is an unlimited
amount of air, hydrogen is always the limiting reagent. The balanced equation is
as follows:
N2 (g) +
3 H2 (g)

2 NH3 (g)
If 600.0 g of hydrogen gas were consumed in this reaction, how many grams of
ammonia (NH3) would be produced? How many grams of nitrogen gas would be
consumed?
H2 is limiting, so: g H2→ moles H2 → moles NH3 → g NH3
600.0 g H2 x (1 mole/ 2.016 g) x (2 NH3/ 3 H2) x 17.03 g/mole = 3379 g of NH3
600.0 g H2 x (1 mole/ 2.016 g) x (1 N2/ 3 H2) x 28.01 g/mole = 2779 g of N2
OR...
3379 g NH3 – 600.0 g H2 = 2779 g N2 (conservation of mass in balanced eqn)
4. 2 Ag2CO3

4 Ag (s) +
2 CO2 (g)
+
O2 (g)
Is this a redox reaction? Why or why not? If you start with 55.9 grams of
silver carbonate and your % yield of silver is 73.9%, how much silver do you
recover?
oxidation numbers
Ag +1, C +4, O →
Ag 0, C +4, O in CO2 is -2, O in O2 is 0
since oxidation numbers on silver and oxygen change, this is a redox rxn
theoretical yield: g Ag2CO3→ moles Ag2CO3→moles Ag→grams Ag
55.9 g Ag2CO3 x (1 mole/275.7) x 4 Ag/2 Ag2CO3 x 107.9 g/mole = 43.8 g
silver
% yield = actual/theoretical, so actual = % x theoretical
73.9% of 43.8 = 32.3 g (remember not to use rounded off numbers in a
calculation!)
5. In the reaction below, 14.0 grams of propane (C3H8) are burned in the presence of
38.9 g of oxygen. Which is the limiting reagent? How many grams of water is
produced? How many grams of carbon dioxide is produced?
1 C3H8 (g) +
5 O2(g)
→
4 H2O(g)
+
3 CO2(g)
find limiting reagent (g propane → moles propane → moles oxygen → g oxygen,
or you could do g oxygen → moles oxygen → moles propane → g propane, either
will give the same answer to which is limiting)
14.0 g C3H8 x (1 mole/44.09g) x 5 O2/ 1 C3H8 x 32.00 g/mole = 50.8 g O2
since we don’t have 50 g of oxygen, oxygen is limiting. product must be
calculated using limiting reagent.
38.9 g O2 x 1 mole/ 32.00g x 4 H2O/5 O2 x 18.02g/mole = 17.5 g water
38.9 g O2 x 1 mole/ 32.00g x 3 CO2/5 O2 x 46.01g/mole = 33.6 g CO2