Worksheet Chapter 5 answers

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					                        Chemistry 105 Worksheet Chapter 6

1.   Balance the following equations.
     2 Mg (s)       +       2 H2O (l)             2 Mg(OH)2 (aq)        + 1     H2(g)

     10 Na (g) +    2 KNO3 (s)  1 K2O (s)+        5 Na2O (s)     +      1 N2 (g)

     2 NaHCO3 (s)          1 Na2O(s)        +     2 CO2 (g)+     1 H2O (g)

     2 C4H10(g)     +13 O2(g)               8 CO2(g)             +      10 H2O(g)

     1 Li2CO3 (aq) +        2 HCl (aq)  1 CO2 (g) +       1 H2O (l) +   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

				
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