Question

Potassium superoxide, KO2, is often used in oxygen masks (such as those used by firefighters) because KO2 reacts with CO2 to release molecular oxygen. Experiments indicate that 2 mol of KO2(s) react with each mole of CO2(g).

(a) The products of the reaction are K2CO3(s) and O2(g). Write a balanced equation for the reaction between KO2(s) and CO2(g).
(b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?
(c) What mass of KO2(s) is needed to consume 18.0 g CO2(g)? What mass of O2(g) is produced during this reaction?

Answer 1

a:
`4KO_2(s) +2CO_2(g) = 2K_2CO_3(s) +3O_2(g)`

b: Oxygen is the only element which is getting oxidised as well reduced

c:
`mass of KO_2 =58.22 gram` and
`mass of O_2 =19.68 gram`

Step-by-step explanation:

Part a: Balanced equation for the given reaction;

Rule:

step1 : first balance the metal

step2: Non metal except oxygen

step 3: lastly balance the oxygen

`4KO_2(s) +2CO_2(g) = 2K_2CO_3(s) +3O_2(g)`

Part b: indicating oxidation number of each element both side

reactant side oxidation number:

K = +1

O= -0.5 and -2

C =+4

Product side oxidation number:

K = +1

O= 0 and -2

C =+4

From the above data it clearly that Oxygen is the only element which is getting oxidised as well reduced .

Part c: Mass calculation

from the balance equation it is clearly that 4 moles of
`KO_2` need 2 moles of
`CO_2` for complete reaction i.e. mole
`KO_2` is used double the mole of
`CO_2`

`Mole of CO2 = 18/44 mol`

`Mole of CO2 = 0.41 mol`

hence the mole of
`KO_2` is two times of
`CO_2`

`Mole of KO_2 = 0.82 mol`

`Mass of KO_2 = mole * molecular weight`

`mass of KO_2 =0.82 *71 gram`

`mass of KO_2 =58.22 gram`

From 2 mole of
`CO_2` 3 mole of
`O_2` is produced

From 1 mole of
`CO_2` 1.5 mole of
`O_2` is produced

From 0.41 mole of
`CO_2`
`1.5*0.41` mole of
`O_2` is produced

`Mass of O_2 = mole * molecular weight`

`mass of O_2 =1.5 *0.41* 32 gram`

`mass of O_2 =19.68 gram`