1. C 2 H 2 (g) + 5/2 O 2 (g) à 2 CO 2 (g) + H 2 O(l) H = -1299.5 kJ C(s) + O 2 (

Question

1. C2H2(g) + 5/2 O2(g) à 2 CO2(g) + H2O(l)                                 H = -1299.5 kJ

            C(s) + O2(g) à CO2(g)                                                      H = -393.5 kJ

            H2 + ½ O2(g) à H2O(l)                                                     H = -285.8 kJ

Calculate Hrxnfor the following reaction:

    2 C(s) + H2(g) à C2H2(g)                                                  H = ???

2.

Oxidation is a common reaction for many metals. Given the following equations, first balance them, then determine which species are oxidized and which are reduced. Then look at the “activity series” table in your textbook and determine whether the reaction can happen or not.

            a)   Fe(s) +   Ag2SO4(aq) à FeSO4(aq) + Ag(s)

            b)   Mg(NO3)2(aq) + Li (s) à LiNO3(aq) + Mg(s)

            c)    Cu(MnO4)2(aq) + K(s) à KMnO4(aq) + Cu(s)

3. The reaction between potassium superoxide, KO2, and CO2is used as a source of O2and absorber of CO2in self-contained breathing equipment used by rescue workers. A) If 40.0 grams of KO2and 40.0 grams of CO2are present in the container, how many grams of O2are produced? B) Calculate the percent yield if 10 g of O2were actually used until the tank was empty.

4 KO2 + 2 CO2 à 2 K2CO3 + 3 O2

Explanation / Answer

1.

(a)        C2H2(g) + 5/2 O2(g) = 2 CO2(g) + H2O(l)                   H = -1299.5 kJ

(b)       2C(s) + 2O2(g) = 2CO2(g) H = -393.5*2 kJ

(c)       H2 + ½ O2(g) = H2O(l) H = -285.8 kJ

Let us consider the following equation as (d)         

(d)       2 C(s) + H2(g) = C2H2(g)                                                  H = ???

Hrxn for the above reaction can be calculated as following:

Eq (d) = eq [2(b) +(c)] - eq (a)

So, H = (-393.5*2-285.8) – (-1299.5)

            = 226.7 kJ

2.

Balanced redox reaction.

            a)   Fe(s) +   Ag2SO4(aq) = FeSO4(aq) + 2Ag(s)

Fe (from 0 to +2) oxideized and Ag+ (from +1 to 0) reduced.(Possible, Fe is more reactive)

Balanced redox reaction.

            b)   Mg(NO3)2(aq) + 2Li (s) = 2LiNO3(aq) + Mg(s)

Mg2+ (from +2 to 0) reduced and Li (from 0 to +1) oxidized. (Possible, Li is more reactive)

Balanced redox reaction.

            c)    Cu(MnO4)2(aq) + 2K(s) = 2KMnO4(aq) + Cu(s)

Cu2+ (from +2 to 0) reduced and K (from 0 to +1) oxidized (Possible, K is more reactive)

3.

4 KO2 + 2 CO2 = 2 K2CO3 + 3 O2

(A)       4 moles of KO2 produces 3 moles of O2

            Molar mass of KO2 = 71.1 g/mol

            Molar mass of CO2 = 44 g/mol

40 g of KO2 = 40/71.1 moles = 0.563 moles.

40 g of CO2 = 40/44 moles =0.91 moles.

So, KO2 is the limiting reagent.

Thus 0.563 moles of KO2 will produce = 0.563*3/4 moles of O2

In grams, the weight of O2 produced = 32*0.563*3/4 = 13.5 g

(B)Theoretical yield = 13.5 g. Actual yield = 10 g

So, Yield = 100*10/13 = 76.92%

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